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Book 






PRESENTED BY 



I 



WENTWORTH'S 
SERIES OF MATHEMATICS. 



First Steps in Number. 

Primary Arithmetic. 

Grammar School Arithmetic. 

High School Arithmetic. 

Exercises in Arithmetic. 

Shorter Course in Algebra. 

Elements of Algebra. Complete Algebra. 

College Algebra. Exercises in Algebra. 

Plane Geometry. 

Plane and Solid Geometry. 

Exercises in Geometry. 

PI. and Sol. Geometry and PI. Trigonometry* 

Plane Trigonometry and Tables. 

Plane and Spherical Trigonometry. 

Surveying. 

PI. and Sph. Trigonometry, Surveying, and Tables. 

Trigonometry, Surveying, and Navigation. 

Trigonometry Formulas. 

Logarithmic and Trigonometric Tables (Seven). 

Log. and Trig. Tables (Complete Edition). 

Analytic Geometry. 



Special Terms and Circular on Application. 



PLANE AND SPHERICAL 



TRIGONOMETRY. 



BY 

G. A. WENTWORTH, A.M., 

PROFESSOR OF MATHEMATICS IN PHILLIPS EXETER ACADEMY. 



>J*«c 



BOSTON: 

PUBLISHED BY GINN & COMPANY. 

1887. 



QAs^i 

•VI 4* 85* 



Entered, according to Act of Congress, in the year 1882, by 

G. A. WENTWORTH, 

in the office of the Librarian of Congress, at Washington 

Gift from 
Mr. James F'cKSrdy 

Jr.r-. 19,1932 



J. S. Cvshinq & Co., Printers, Boston. 



PREFACE. 



IN preparing this work the aim has been to famish just so much 
of Trigonometry as is actually taught in our best schools and 
colleges. Consequently, development of functions in series and all 
other investigations that are important only for the special student 
have been omitted. " The principles have been unfolded with the 
utmost brevity consistent with simplicity and clearness, and inter- 
esting problems have been selected with a view to awaken a real 
love for the study. Much time and labor have been spent in devis- 
ing the simplest proofs for the propositions, and in exhibiting the 
best methods of arranging the logarithmic work. 

The object of the work on Surveying and Navigation is to pre- 
sent these subjects in a clear and intelligible way, according to the 
best methods in actual use ; and also to present them in so small a 
compass that students in general may find the time to acquire a 
competent knowledge of these very interesting and important studies. 

The author is under particular obligation for assistance to G. A. Hill, 
A.M., of Cambridge, Mass., and to Prof. James L. Patterson, of Law- 
renceville, N.J. 

G. A. WENTWORTH. 
Phillips Exeter Academy, 
September, 1882. 



CONTENTS, 



PLANE TRIGONOMETRY. 

CHAPTER I. Functions of Acute Angles: 

Definitions, 1 ; representation of functions by lines, 7 ; changes in 
the functions as the angle changes, 9 ; functions of complementary 
angles, 10 ; relations of the functions of an angle, 11 ; formulas for 
finding all the other functions of an angle when one function of the 
angle is given, 13 ; functions of 45°, 30°, 60°, 15. 

CHAPTER II. The Right Triangle: 

Solution : Case I., when an acute angle and the hypotenuse are 
given, 16 ; Case II., when an acute angle and the opposite leg are 
given, 17; Case III., when an acute angle and the adjacent leg 
are given, 17 ; Case IV., when the hypotenuse and a leg are given, 
18 ; Case V., when the two legs are given, 18 ; general method of 
solving the right triangle, 19 ; area of the right triangle, 20 ; the 
isosceles triangle, 24 ; the regular polygon, 26. 

CHAPTER III. Goniometry: 

Definition of Goniometry, 28 ; angles of any magnitude, 28 ; gen- 
eral definitions of the functions of angles, 29; algebraic signs of the 
functions, 31 ; functions of a variable angle, 32 ; functions of angles 
larger than 360°, 34 ; formulas for acute angles extended to all angles, 
35 ; reduction of the functions of all angles to the functions of angles 
in the first quadrant, 38 ; functions of angles that differ by 90°, 40; 
functions of a negative angle, 41 ; functions of the sum of two angles, 
43 ; functions of the difference of two angles, 45 ; functions of twice 
an angle, 47 ; functions of half an angle, 47 ; sums and differences of 
functions, 48. 



VI TRIGONOMETRY. 



CHAPTER IV. The Oblique Triangle: 

Law of sines, 50 ; law of cosines, 52 ; law of tangents, 52. Solu- 
tion : Case I., when one side and two angles are given, 54 ; Case II., 
when two sides and the angle opposite to one of them are given, 56 ; 
Case III., when two sides and the included angle are given, 60 ; Case 
IV., when the three sides are given, 64 ; area of a triangle, 68 ; mis- 
cellaneous problems, 71-88. 

EXAMINATION PAPERS, 89-102. 



SPHERICAL TRIGONOMETRY. 

CHAPTER V. The Right Spherical Triangle : 

Introduction, 103 ; formulas relating to right spherical triangles, 
105 ; Napier's rules, 108. Solution : Case I., when the two legs are 
given, 110 ; Case II., when the hypotenuse and a leg are given, 110 ; 
Case III., when a leg and the opposite angle are given, 111 ; Case IV., 
when a leg and an adjacent angle are given, 111 ; Case V., when the 
hypotenuse and an oblique angle are given, 111 ; Case VI., when 
the two oblique angles are given, 111 ; solution of the isosceles 
spherical triangle, 116 ; solution of a regular spherical polygon, 116. 

CHAPTER VI. The Oblique Spherical Triangle : 

Fundamental formulas, 117 ; formulas for half angles and sides, 
119 ; Gauss's equations and Napier's analogies, 121. Solution : Case I., 
when two sides and the included angle are given, 123 ; Case II., when 
two angles and the included side are given, 125 ; Case III., when two 
sides and an angle opposite to one of them are given, 127 ; Case IV., 
when two angles and a side opposite to one of them are given, 129 ; 
Case V., when the three sides are given, 130; Case VI., when the 
three angles are given, 131 ; area of a spherical triangle, 133. 

CHAPTER VII. Applications of Spherical Trigonometry : 

Problem, to reduce an angle measured in space to the horizon, 136 ; 
problem, to find the distance between two places on the earth's sur- 
face when the latitudes of the places and the difference of their longi- 
tudes are known, 137 ; the celestial sphere, 137 ; spherical co-ordinates, 
140 ; the astronomical triangle, 142; astronomical problems, 143-146. 



PLANE TBIGONOMETKY. 



CHAPTEE I. 



TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES. 



§ 1. Definitions. 

The sides and angles of a plane triangle are so related that 
any three given parts, provided at least one of thern is a side, 
determine the shape and the size of the triangle. 

G-eometry shows how, from three such parts, to construct the 
triangle and find the values of the unknown parts. 

Trigonometry shows how to compute the unknown parts of a 
triangle from the numerical values of the given parts. 

Geometry shows in a general way that the sides and angles 
of a triangle are mutually dependent. Trigonometry begins 
by showing the exact nature of this dependence in the right 
triangle, and for this purpose employs the ratios of its sides. 

Let MAN (Fig. 1) be an acute angle. If from any points 

B, D, F, in one of its sides 

perpendiculars BC, BE, FG, 

are let fall to the other side, then 
the right triangles ABC, ABE, 

AFG, thus formed have the 

angle A common, and are there- 
fore mutually equiangular and 
similar. Hence, the ratios of 
their corresponding sides, pair by 
pair, are equal. That is, 




AC _AE ^AG , 
AB AD AF' 



AC == AE = AG 
BC BE FG 



; etc. 



TRIGONOMETRY. 



Hence, for every value of an acute angle A there are certain 
numbers that express the values of the ratios of the sides in 
all right triangles that have this acute angle A. 
There are altogether six different ratios : 

I. The ratio of the opposite leg to the hypotenuse is called 
the Sine of A, and is written sin A. 

II. The ratio of the adjacent leg to the hypotenuse is called 
the Cosine of A, and written cos -4. 

III. The ratio of the opposite leg to the adjacent leg is 
called the Tangent of A, and written tan A. 

IV. The ratio of the adjacent leg to the opposite leg is 
called the Cotangent of A, and written cot .4. 

V. The ratio of the hypotenuse to the adjacent leg is called 
the Secant of A, and written sec ^4. 

VI. The ratio of the hypotenuse to the opposite leg is called 
the Cosecant of A, and written cscA. 




In the right triangle ABC 
(Fig. 2) let a, b, c denote the 
lengths of the sides opposite to 
the acute angles A, B, and the 
right angle C, respectively, these 
lengths being all expressed in 
terms of a common unit. Then, 



cin ^_fl_ °PP ositele g 
c hypotenuse 

_ a _ opposite leg 



tan A 



sec A 



cos A 



adjacent leg 
hypotenuse 



b adjacent leg 

c hypotenuse \ 

b adjacent leg 



^ ot ^ = J = adjacentleg 
a opposite leg 



esc A 



c hypotenuse 

a opposite leg 



These six ratios are called the Trigonometric Functions of the 
angle A. 



TRIGONOMETRIC FUNCTIONS. 



Exercise I. 

1. What are the functions of the other acute angle B of the 
triangle ABC (Fig. 2)? 

2. Prove that if two angles, A and B, are complements of 
each other (i.e., i£A + B = 90°), then, 

sin A = cos B, tan A = cot B, sec A = esc B ; 
cos A = sin i?, cot A = tan i?, esc A = sec i?. 

3. Finot the values of the functions of A, if a, b } c respec- 
tively have the following values : 

(i.) 3, 4, 5. (iii.) 8, 15, 17. (v.) 3.9, 8, 8.9. 
(ii.) 5, 12, 13. (iv.) 9, 40, 41. (vi.) 1.19, 1.20, 1.69. 

4. What condition must be fulfilled by the lengths of the 
three lines a, b } c (Fig. 2) in order to make them the sides of 
a right triangle ? Is this condition fulfilled in Example 3 ? 

5. Find the values of the functions of A, if a, b, c respec- 
tively have the following values : 

(i.) 2mn, m 2 — n 2 , m 2 -\-n 2 . (iii.) pqr, qrs, rsp. 

(n.) £_, x + y, — i -^-. (iv.) — , — , — . 

x — y X ~V pq sq ps 

6. Prove that the values of a, b, c, in (i.) and (ii.), Example 
5, satisfy the condition necessary to make them the sides of 
a right triangle. 

7. What equations of condition must be satisfied by the 
values of a, b, c, in (iii.) and (iv.), Example 5, in order that the 
values may represent the sides of a right triangle ? 

Compute the functions of A and B when, 



8. a = 24, Z> = 143. 11. a=-Vp 2 + q 2 , b = V2pq. 

9. a = 0.264, c = 0.265. 12. a = ^/p 2 +pq, c = p + q. 
10. 5 = 9.5, , = 19.3. is 6 = 2V^, c=p + q . 



6 TRIGONOMETRY. 




Compute the functions of A when, 




14. a = 2b. 16. a + 5 = 


= f*. 


15. a = fc. 17. a — b = 


4? 


18. Find a if sb A = f and c = 20.5. 




19. Find b if cos J. = 0.44 and c = 3.5. 




20. Find a if tan ^4 = ^ and 5 = 2-^-. 




21. Find 6 if cot^4 = 4 and a = 17. 




22. Find c if sec A = 2 and b = 20. 





23. Find c if esc A = 6.45 and a = 35.6. 
Construct a right triangle : given, 

24. c = 6, tan^4 = f. 26. 5 = 2, sin^4 = 0.6. 

25. a=3.5, cos^4=|. 27. 5 = 4, esc A = 4. 

28. In a right triangle, c = 2.5 miles, sin ^4 = 0.6, cos ^4 = 
0.8 ; compute the legs. 

29. Construct (with a protractor) the A 20°, 40°, and 70°; 
determine their functions by measuring the necessary lines, 
and compare the values obtained in this way with the more 
correct values given in the following table : 



20° 
40° 
70° 


sin 


cos 


tan 


cot 


sec 


CSC 


0.342 
0.643 
0.940 


0.940 
0.766 
0.342 


0.364 
0.839 

2.747 


2.747 
1.192 
0.364 


1.064 
1.305 
2.924 


2.924 
1.556 
1.064 



30. Find, by means of the above table, the legs of a right 
triangle if A = 20°, c = 1; also, if A = 20°, c = 4. 

31. In a right triangle, given a = 3 and c = 5; find the 
hypotenuse of a similar triangle in which a = 240,000 miles. 

32. By dividing the length of a vertical rod by the length 
of its horizontal shadow, the tangent of the angle of elevation 
of the sun at the time of observation was found to be 0.82. 
How high is a tower, if the length of its horizontal shadow at 
the same time is 174.3 yards? 



TRIGONOMETRIC FUNCTIONS. 




§ 2. Representation of Functions by Lines. 

The functions of an angle, being ratios, are numbers; but 
we may represent them by lines if we first choose a unit of 
length, and then construct right triangles, such that the de- 
nominators of the ratios shall be equal to this unit. The most 
convenient way to do this is as follows : 

About a point (Fig. 3) as a 
centre, with a radius equal to one 
unit of length, describe a circle 
and draw two diameters A A' and 
BB' perpendicular to each other. 

The circle with radius equal to 
1 is called a unit circle, AA! the 
horizontal, and BB' the vertical 
diameter. 

Let A OP be an acute angle, 
and let its value (in degrees, etc.) 
be denoted by x. We may regard 

the Z. x as generated by a radius OP that revolves about 
from the position OA to the position shown in the figure ; 
viewed in this way, OP is called the moving radius. 

Draw PM J- to OA. In the rt. A OPM the hypotenuse 
OP= 1 ; therefore, sin a; = PM; cosrr == OM. 

Since PM is equal to ON, and OiVis the projection of OP 
on BB', and since OM is the projection of OP on AA1 , there- 
fore, in a unit circle, 

sin a; == projection of moving radius on vertical diameter ; 

cos a; = projection of moving radius on horizontal diameter. 

Through A and B draw tangents to the circle meeting OP, 
produced in T and 8, respectively ; then, in the rt. A OA T, 
the leg OA = l, and in the rt. A OB 8, the leg 0JB = 1; while 
the Z 08B = Z.x (why ?). Therefore, 

imx = AT) sec# = OT; cotx = B8; cscx = 0S, 



8 TRIGONOMETRY. 



These six line values (as they may be termed) of the func- 
tions are all expressed in terms of the radius of the circle as a 
unit ; and it is clear that as the angle varies in value the line 
values of the functions will always remain equal numerically 
to the ratio values. Hence, in studying the changes in the 
functions as the angle is supposed to vary, we may employ the 
simpler line values instead of the ratio values. 

Exercise II. 

1. Eepresent by lines the functions of a larger angle than 
that shown in Fig. 3. 

2. Show that sin x is less than tana;. 

3. Show that sec a; is greater than tana;. 

4. Show that esc x is greater than cota;. 
Construct the angle x if, 

5. tana; = 3. 7. cosa; = -|-. 9. sina; = 2cosa;. 

6. esc a; = 2. 8. sina; = cosa;. 10. 4sina; = tana;. 

11. Show that the sine of an angle is equal to one-half the 
chord of twice the angle. 

12. Find x if sin a; is equal to one-half the side of a regular 
inscribed decagon. 

13. Given x and y {x-\-y being less than 90°) ; construct the 
value of sin {x -f- y) — sin a;. 

14. Given x and y (x-\~y being less than 90°) ; construct the 
value of tan (x-\-y) — sin (x + y) -f tana; — sin a;. 

Given an angle x ; construct an angle y such that, 

15. siny = 2sina;. 17. tany = 3tana;. 

16. cosy = -J cos x. 18. secy = csca;. 

19. Show by construction that 2 sin A > sin 2 A. 

20. Given two angles A and B ( A -f- B being less than 90°) , 
show that sin ( A -f B) < sin A -f- sin B. 

21. Given sin a; in a unit circle; find the length of a line 
corresponding in position to sin a; in a circle whose radius is r ? 

22. In a right triangle, given the hypotenuse c, and also 
sin A = m, cos A = n ; find the legs. 



TRIGONOMETRIC FUNCTIONS. 




§ 3. CHANGES IN THE FUNCTIONS AS THE ANGLE CHANGES. 

If we suppose the A AOP, or x (Fig. 4) to increase gradu^ 
ally by the revolution of the moving 
radius OP about 0, the point P will 
move along the arc AB towards B, 
T will move along the tangent A T 
away from A, S will move along 
the tangent BS towards B, and M 
will move along the radius OA 
towards 0. 

Hence., the lines PM, AT, OT 
will gradually increase in length, 
and the lines OM, BS, OS, will 
gradually decrease. That is, 

As an acute angle increases, its 
sine, tangent, and secant also in- 
crease, while its cosine, cotangent, and cosecant decrease. 

On the other hand, if we suppose x to decrease gradually, 
the reverse changes in its functions will occur. 

If we suppose x to decrease to 0°, OP will coincide with OA 
and be parallel to BS. Therefore, PM and A T will vanish, 
OMwill become equal to OA, while BS &no\ OS will each be 
infinitely long, and be represented in value by the symbol oo. 

And if we suppose x to increase to 90°, OP will coincide 
with OB and be parallel to AT. Therefore, PM and OS will 
each be equal to OB, OM and BS will vanish, while AT and 
Twill each be infinite in length. 

Hence, as the angle x increases from 0° to 90°, 

sin x increases from to 1, 
cos x decreases from 1 to 0, 
tana; increases from to oo, 
cot x decreases from co to 0, 
sec x increases from 1 to oo, 
esc x decreases from oo to 1. 



10 



TRIGONOMETRY. 



The values of the functions of 0° and of 90° are the limiting 
values of the functions of an acute angle. It is evident that 
(disregarding the limiting values), 

Sines and cosines are always less than 1 ; 

Secants and cosecants are always greater than 1 ; 

Tangents and cotangents have all values between and oo. 

Remark. ^Ve are now able to understand why the sine, cosine, etc., 
of an angle are called junctions of the angle. By a function of any mag- 
nitude is meant another magnitude which remains the same so long as 
the first magnitude remains the same, but changes in value for every 
change in the value of the first magnitude. This, as we now see, is the 
relation in which the sine, cosine, eto., of an angle stand to the angle. 




§4. Functions of Complementary Angles. 

The general form of two complementary angles is A and 
90° - A. 

Inthert. A ABC (Fig. 5) 
A + B = 90°, hence B = 90° - A. 
Therefore (§ 1), 

sin A = cos B = cos (90° - A), 
cos A = sin B = sin (90° - A\ 
tan A = cot B = cot (90° - A), 
cot A = tan B = tan (90° - A\ 
sec A = esc B = esc (90° — A), 
esc A = sec B = sec (90° — A). 

Therefore, 

Each function of an acute angle is equal to the co-named 
function of the complementary angle. 

Note. Cbsine, cotangent, and cosecant are sometimes called co-functions ; 
the words are simply abbreviated forms of complement's sine, complement's 
tangent, and complement 1 s secant. 

Hence, also, 

Any function of an angle between 45° and 90° may be found 
by talcing the co-named function of the complementary angle 
between 0° and 4:0°. 



TRIGONOMETRIC FUNCTIONS. 11 



Exercise III. 

1. Express the following functions as functions of the com- 
plementary angle : 

sin 30°. tan 89°. esc 18° 10'. cot 82° 19'. 
cos 45°. cot 15°. cos 37° 24'. esc 54° 46'. 

2. Express the following functions as functions of an angle 
less than 45° : 

sin 60°. tan57°. csc69°2'. cot 89° 59'. 

cos 75°. cot 84°. cos85°39'. esc 45° 1'. 

3. Given tan 30° = i V3 ; find cot 60°. 

4. Given tan A = cot A ; find A. 

5. Given cos A = sin 2 A ; find A. 

6. Given sin A = cos 2 A ; find A. 

7. Given cos A = sin (45° — J A) ; find A. 

8. Given cot $ A = tan A ; find A. 

9. Given tan (45° -f A) = cot A ; find ^4. 

10. Find J. if sin A = cos 4 J.. 

11. Find ^4 if cot A = tan 8 A 

12. Find A if cot A = tan rc J.. 

el e 

§ 5. Relations of the Functions of an Angle. 



Since (Fig. 5) a 2 -{-b 2 = c 2 , therefore, 

a 2 . b 2 _ AA 2 , /A 2 , 

Therefore (§ 1), (sin ^4) 2 + (cos ^4) 2 = 1 ; 

or, as usually written for convenience, 

sin 2 A + cos 2 A = l. [1] 

That is : The sum of the squares of the sine and the cosine of 
an angle is equal to unity. 



12 TRIGONOMETRY. 



Formula [1] enables us to find the cosine of an angle when 
the sine is known, and vice versa. The values of sin A and of 
cos A deduced from [1] are : 



sin 



A = .Vl — cosM., cos A = Vl — sin 2 A. 



a . abaca 

Since — \ — = _X7 = T , 

c c c b o 

sin A ro -i 

therefore (§ 1), tan A = ^j- MJ 

That is : The tangent of an angle is equal to the sine divided 
by the cosine. 

Formula [2] enables us to find the tangent of an angle when 
the sine and the cosine are known. 

a- a c ~ be.. , a „ b ~ 

Since -x- = l, -X 7 = l, and T X- = 1, 
c a cb o a 

therefore (§1), sin A X esc A = 1 1 

cos A x sec A = 1 \ • [3] 

tan A x cot A = 1 J 

That is : The sine and the cosecant of an angle, the cosine 
and secant, and the tangent and cotangent, pair by pair, are 
reciprocals. 

The equations in [3] enable us to find an unknown function 
contained in any pair of these reciprocals when the other func- 
tion in this pair is known. 

Exercise IV. 

1. Prove Formulas [1] - [3], using for the functions the line 
values in unit circle given in § 3. 

2. Prove that 1 -f- tan 2 J. = secM. 

Hint. Divide the terms of the equation a 2 + & 8 = c 2 by 6 s . 

3. Prove that 1 + cot 2 A = esc 2 A. 

cos A 

4. Prove that cot A = - — j* 

ma A 



TRIGONOMETRIC FUNCTIONS. 13 

§ 6. Application of Formulas [1J - [3]. 

Formulas [1], [2], and [3] enable us, when any one function 
of an angle is given, to find all the others. A given value of 
any one function, therefore, determines all the others. 

Example 1. Given sin A = f ; find the other functions. 

By [1], cos A = VT=$ = V| = i V5. 

By[2], tan^f^vH^TB 

By [3], cotA = ^I, secA = —, csc^ = -- 

Example 2. Given tan A = 3 ; find the other functions. 

By [2], »™4 = S. 

cos A 

And by [1], sin 2 J. -f- cos 2 A = 1. 

If we solve these equations (regarding sin A and cos A as 
two unknown quantities), we find that, 

sin A = 3 V T V cos A — V^o • 
Then by [3], cot A=%, secA = VIO, esc A = -jVlO. 

Example 3. Given sec A = m ; find the other functions. 
1 



By [3], cos A = 



m 



By[l], B in^ = Jl-l = A P3 = I V ^Tl. 

By [2], [3], tan A = Vri^l , cot A = = 



Vm 2 



esc A = , 



14 TRIGONOMETRY. 



Exercise V. 
Find the values of the other functions when, 

1. sinJ. = f|. 5. tanJ. = f. 9. csc^4 = V2 

2. sin A = 0.8. 6. cot A = 1. 10. sin^L = m. 

3. co8A = $$. 7. cot .4 = 0.5. 11. sin A = 



2m 



1 + m 2 

4. cos.4 = 0.28. 8. sec A = 2. 2mn 

12. cos ^4 = -— — ■ 
rrr-f-nr 

13. Given tan 45° = 1 ; find the other functions of 45°. 

14. Given sin 30° = -J ; find the other functions of 30°. 

15. Given esc 60° = -§V3 ; find the other functions of 60°. 

16. Given tan 15° = 2 — V3 ; find the other functions of 15°. 

17. Given cot 22° 30'= V2 + 1; find the other functions 
of 22° 30'. 

18. Given sin 0° = ; find the other functions of 0°. 

19. Given sin 90° = 1 ; find the other functions of 90°. 

20. Given tan 90° = oo ; find the other functions of 90°. 

21. Express the values of all the other functions in terms 
of sin A. 

22. Express the values of all the other functions in terms 
of cos A. 

23. Express the values of all the other functions in terms 
*}f tan A. 

24. Express the values of all the other functions in terms 
of cot A. 

25. Given 2 sin A = cos .4 ; find sin A and cos A. 

26. Given 4 sin A = tan A ; find sin A and tan A. 

27. If sin A : cos ^4 = 9 : 40, find sin A and cos A, 

28. Transform the quantity tan 2 J. -f cot 2 A — sin 2 A — cos 2 A 
into a form containing only cos A. 

29. Prove that sin A -f cos A = (1 + tan A) cos A. 

30. Prove that tan A -f cot .4 = sec A X esc .4. 



TRIGONOMETRIC FUNCTIONS. 



15 



1 



§ 7. Functions of 45°. 

Let ABC (Fig. 6) be an isosceles right triangle, in which 
the length of the hypotenuse AB 
is equal to 1 ; then A C is equal to 
BO, and the angle A is equal to 45°. 
Since AC 2 -f ~W? = 1, _ therefore 
2AC 2 = 1, and^C = Vj = lV2. 

Therefore (§ 1), 



sin 45° 


= cos 45° 


= 1V2 


tan 45° 


= cot 45° 


= 1. 


sec 45° 


= esc 45° 


= V2. 




§8. Functions of 30° and 60°. 

Let ABC be an equilateral triangle in which the length 
of each side is equal to 1 ; and let CD bisect the angle C. 
Then CD is perpendicular to AB and bisects AB; hence, 
AD = I and CD = VT~T= Vf = lV5. 

In the right triangle J. DC, the angle ACD = S0°, and the 
angle CAD = 60°. 

"Whence (§ 1), 



sin 30° = cos 60° = J. 



cos 30° = sin 60° 
tan 30° = cot 60° = 

cot 30° = tan 60° = 
sec 30° = esc 60° = 



iV3. 
J_ = 

V3 

V3. 
2 



W.3- 



-#Va 




V3 
esc 30° — sec G0° = 2. 

The results for sine and cosine of 30°, 45°, and 60° may be 
easily remembered by arranging them in the following form : 



Angle . . . 


30° 


45° 


60° 


£ = 0.5 


Sine .... 


h 


*V2 


*V3 


*V2 = 0.70711 


Cosine. . . 


iV3 


*V2 


I 


£V3 = 0.86603 






/ 


> 





OSAPTM II. 



THE RIGHT TRIANGLE. 



§ 9. The Given Parts. 

In order to solve a right triangle, two parts besides the right 
angle must be given, one of them at least being a side. 
The two given parts may be : 

I. An acute angle and the hypotenuse. 
II. An acute angle and the opposite leg. 

III. An acute angle and the adjacent leg. 

IV. The hypotenuse and a leg. 
V. The two legs. 

§ 10. Case I. 
Given A = 34° 28' and c = 18.75 ; required B, a, b. 



1. .5=90°- ^ = 55° 32'. 




2. - =sin^4; .". a = csinA. 



b 

Fig. 8. 

log a = log C + log sin A 
\ogc = 1.27300 
log sin A = 9.75276* 

loga = 1.02576 
a = 10.61 



3. - =cosJ.; .'. 6 = ccos^4. 



log b = log c -f- log cos A 
\ogc = 1.27300 
log cos A= 9.91617 



\ogb = 1.18917 
b t= 15.459 



* For Logarithms, and directions how to use thein, see Wentworth 
and Hill's Five-place Tables. 

When —10 belongs to a logarithm or cologarithm, and is not written, 
it must be remembered that the logarithm or cologarithm is 10 too large, 



THE RIGHT TRIANGLE. 



17 



~-> 



§ 11. Case II. 
Given A = 62* 10 r , a = 78 ; find J?, 5, <?. 



1. ^ = 90°-^ = 27° 50'. 

2. - =cot^4.; .'. 5 = a cot ^4. 

3. — = sin A. 
c 

.'. a =csmA, and c = - 



sin ^4 




log& = log a -f log cot A 
log a = 1.89209 
log cot A = 9.72262 

log b = 1.61471 
b =41.182 



logc = loga-f-cologsin^4 

loga = 1.89209 

cologsin^4= 0.05340 

loge = 1.94549 

c = 88.204 



§ 12. Case III. 
Given ^ = 50° 2', 6 = 88 ; find B, a, c. 

1. 5 = 90° - ^4 = 39° 58'. 

2. 7 =tan^4; .'.a = bta,nA. 
b 

3. - =cos^4. 
c 

:. b =<?cos^4, and c 



cos A 



log a = log b -f- log tan J. 
log J = 1.94448 
log tan A= 10.0767 



log a = 2.02118 
a = 105.0 




logc = log£~f-cologcosvl 

logS = 1.94448 

cologcosJ= 0.19223 

logc = 2.13671 

c = 137.0 



18 



TRIGONOMETRY. 





§13. 


Case IV. 

Given c = 58.40, a = 47.55 ; find 




B 


A, B, b. 


CyS 


a 


1 ' A a 

1. sin .4 = -• 
c 


* / 


C 


2. £ = 9QP-A. 


A b 

Kg. IL 


3. - =cotJ.; .-. b = acotA. 
a 


log sin A = log a -f- cole 


g° 


log b = log a + log cot ^4 


log a =1.67715 




loga = 1.67715 


eologe =8.23359 




log cot A= 9.85300 


log sin ^4 = 9.91074 


log& = 1.53015 


A = 54° 31' 




b = 33.896 


B = 35° 29' 








§14. 


Case V. 

Given a = 40, 6 = 27; find A, 


> 


5 


J5 y c. 


CyS 




1. tan^ = ^- 


/ 


a 


2. 5 = 90° -A 




C 


3. - = sinA 


A b 


c 


Kg. 12. 




n — p c i n 4 • * /* — 




sin A 


log tan J. = log a -f col 


3gb 


loge =logrt+cologsin^4 


log a = 1.60206 




log a = 1.60206 


colog£ = 8.56864 




cologsin^4= 0.08152 


log tan ^ = 10.17070 


loge = 1.68358 


A = 55° 59' 




c =48.259 


^ = 34°1' 







the right triangle. 19 

§ 15. G-eneral Method of Solving the Eight Triangle. 

From these five cases it appears that the general method of 
finding an unknown part in a right triangle is as follows : 

Choose from the equation A-\- B = 90°, and the equations that 
define the functions of the angles, an equation in which the re- 
quired part only is unknown ; solve this equation, if necessary, 
to find the value of the unknown part ; then compute the value, 
using logarithms whenever convenient. 

Note 1. In Cases IV. and V. the unknown side may also be found 
by Geometry, from the equation a 2 + b 2 = c 2 ; whence we obtain 

(for Case IV.) b = VcF^a 2 = y/(c + a)(c~a) ; 



(for Case V.) c = Va 2 + b 2 . 

These equations express the values of b and c directly in terms of the 
two given sides ; and if the values of the sides are simple numbers (e.g. 
5, 12, 13), it is often easier to find b or c in this way. But this value of 
c is not adapted to logarithms, and this value of b is not so readily worked 
out by logarithms as the value of b given in \ 13. 

Note 2. In Case IV. if the given sides (here a and c) are nearly alike 
in value, then A is near 90°, and its value cannot be accurately found 
from the tables, because the sines of large angles differ little in value (as 
u evident from Fig. 4). In this case it is better to find B first, by means 
of a formula proved later (see page 47) ; viz., 

tani^ = A^^; 
ic + a 

and to find b by the method given in Note 1, since the same logarithms 
are used in both cases. 

Example. Given a = 49, c = 50 ; find A, B, b. 



log b = J [log (c~a)+ log (c + a)] 
c — a =1 

c + a =99 

log (c - a) = 0.00000 
log (c + a) = 1.99564 
2 )1.99564 
log b =0.99782 

b =9.95 



log tan |i? = £ [l°g ( c — a ) 

+ colog (c + a)] 

log (c-a) = 0.00000 

colog (c + a) = 8.00436 

2 ) 8.00436 

logtan*J5 =9.00218 

IB =5° 44' 21" 

B =11° 29' 

A =78° 31' 



20 



TRIGONOMETRY. 



§ 16. Area of the Right Triangle. 

It is shown in Geometry that the area of a triangle is equal 
to one-half the product of the base by the altitude. 

Therefore, if a and b denote the legs of a right triangle, and 
F the area, F=iab. 

By means of this formula the area may always be found 
when a and b are given or have been computed. 
For example : Find the area, having given : 



Case I. (§ 10). 


Case IV. (§ 13). 


A = 34° 28', c = 18.75. 


a = 47.54, c = 58.40. 


First find (as in~§ 10) log a 

and log b. 

log (F) =log a+log b -f- colog 2 
loga = 1.02578 
logb = 1.18915 
colog 2 = 9.69897 


First find (as in § 13) log a 

and log b. 

log (F) = log a -f-log b -f- colog 2 
loga =1.67715 
logb =1.53025 
colog 2 =9.69897 


log(F) = 1.91390 
F =82.016 


log(F) = 2.90637 
F = 806.06 



Exercise VI. 

1. In Case II. give another way of finding c, after b has been 
found. 

2. In Case III. give another way of finding c, after a has 
been found. 

3. In Case IV. give another way of finding 5, after the 
angles have been found. 

4. In Case V. give another way of finding c, after the angles 
have been found. 

5. Given B and c ; find A, a, b. 

6. Given B and b ; find A, a, c. 

7. Given B and a ; find A, b, c. 

8. Given b and c ; find A, B, a. 



THE RIGHT TRIANGLE. 



21 



Solve the following triangles : 




9 


Given : 


EEQUIEED : 


a = 6, c = 12. 


A = 30°, B = 60°, 


6 = 10.392. 


10 


J. = 60°, 6 = 4. 


B = 30°, c = 8, 


a = 6.9282. 


11 


A = 30°, a = 3. 


5 = 60°, c = 6, 


6 - 5.1961. 


12 


a = 4, 6 = 4. 


A = .5 = 45°, c = 5.6568. 




13 


a = 2, c = 2.82843. 


^ = £ = 45°, 6 = 2. 




14 


c = 627, J. = 23° 30'. 


5 = 66° 30', a = 250.02, 


6 = 575.0. 


15 


c = 2280, A = 28° 5'. 


5 = 61° 55', a = 1073.3, 


6 = 2011.6. 


-16 


c = 72.15, .A = 39° 34'. 


B =50° 26', a = 45.958, 


6 = 55.620. 


17 


c = l, ^ = 36°. 


j? = 54°, a = 0.58779, 


6 = 0.80902. 


18 


c = 200, J? = 21° 47'. 


^ = 68° 13', a =185.73, 


6 = 74.22. 


19 


c = 93.4, B = 76° 25'. 


^4 = 13° 35', a = 21.936, 


6 = 90.788. 


20 


a = 637, A= 4° 35'. 


B = 85° 25', 6 = 7946, 


c= 7971.5. 


21 


a = 48.532, A = 36° 44'. 


B = 53° 16', 6 = 65.031, 


c = 81.144. 


22 


a = 0.0008, A = 86°. 


j5 = 4°, " 6 = 0.0000559 


, c = 0.000802. 


23 


6 = 50.937, J? = 43° 48'. 


A = 46° 12', a = 53.116, 


c = 73.59. 


24 


6 = 2, B= 3° 38'. 


^i = 86°22', a = 31.497, 


c = 31.560. 


25 


a = 992, 5 = 76° 19'. 


A = 13° 41', 6 = 4074.5. 


c = 4193.6. 


26 


a = 73, 5 = 68° 52'. 


4 = 21° 8, 6 = 188.86, 


c = 202.47. 


27 


a = 2.189, 5 = 45° 25'. 


A = 44° 35', 6 = 2.2211, 


c = 3.1185. 


28 


6 = 4, A = 37° 56'. 


5 = 520 4', a =3.1176, 


c = 5.0714. 


29 


c = 8590, a = 4476. 


A = 31° 24', 5 = 58° 36', 


6 = 7332.8. 


30 


c = 86.53, a =71.78. 


^1 = 56° 3', B = 33° 57', 


6 = 48.324. 


31 


c = 9.35, a = 8.49. 


J. = 65° 14', B = 24° 46', 


6 = 3.917. 


32 


c = 2194, 6=1312.7. 


A = 53° 15', 5 = 36° 45', 


a = 1758. 


33 


c = 30.69, 6 = 18.256. 


A = 53° 30', B = 36° 30', 


a = 24.67. 


34 


a = 38.313, 6 = 19.522. 


A = 63°, 5 = 27°, 


c = 43. 


35 


a = 1.2291, 6 = 14.950. 


A= 4° 42', B = 85° 18', 


c = 15. 


36 


a = 415.38, 6 = 62.080. 


J. = 81° 30', 5= 8° 30', 


c = 420. 


37 


a = 13.690, 6 = 16.926. 


-4 = 38° 58', B = 51° 2', 


c = 21.769. 


38 


c = 91.92, a = 2.19. 


4 = 1°21'55", B = 88°38'5", 6 = 91.894. 


Compute the unknown parts and also the area, having given : 


39. a = 5, b = 6. 44. c = 6S, A- 


= 69° 54'. 


40. a = 0.615, c = 70. 45. c = 27, B- 


= 44° 4'. 


41. b = -\/2, c = V3". 46. a = 47, J5 = 


= 48° 49'. 


42. a =7, 4 = 18° 14'. 47. 5 = 9, B 


= 34° 44'. 


43. 5=12, ^ = 29° 8'. 48. c = K462 £ 


= 86° 4'. 






7 





22 TRIGONOMETRY. 



49. Find the value of F'in terms of c and A. 

50. Find the value of .Fin terms of a and A. 

51. Find the value of .Fin terms of b and A. 

52. Find the value of .Fin terms of a and c. 

53. Given F r = 58, a = 10 ; solve the triangle. 

54. Given F = 18, 5 = 5; solve the triangle. 

55. Given F= 12, J. = 29° ; solve the triangle. 

56. Given F= 100, c = 22; solve the triangle. 

57. Find the angles of a right triangle if the hypotenuse is 
equal to three times one of the legs. 

58. Find the legs of a right triangle if the hypotenuse = 6, 
and one angle is twice the other. 

59. In a right triangle given c, and A = nB ; find a and b. 

60. In a right triangle the difference between the hypote- 
nuse and the greater leg is equal to the difference between the 
two legs ; find the angles. 



The angle of elevation of an object (or angle of depression, 
if the object is below the level of the observer) is the angle 
which a line from the eye to the object makes with a horizon- 
tal line in the same vertical plane. 

61. At a horizontal distance of 120 feet from the foot of a 
steeple, the angle of elevation of the top was found to be 60° 30' ; 
find the height of the steeple. 

62. From the top of a rock that rises vertically 326 feet out 
of the water, the angle of depression of a boat was found to be 
24° ; find the distance of the boat from the foot of the rock. 

63. How far is a monument, in a level plain, from the eye, 
if the height of the monument is 200 feet and the angle of ele- 
vation of the top 3° 30' ? 

64. In order to find the breadth of a river a distance AB 
was measured along the bank, the point A being directly op- 
posite a tree C on the other side. The angle ABCwas also 
measured. If .4^ = 96 feet, and ABC = 21° W, find -the 
breadth of the river. 

If ABC= 45°, what would be the breadth of the river? 



i. 



THE EIGHT TEIANGLE. 23 



e, 



65. Find the angle of elevation of the sun when a tower 
a feet high casts a horizontal shadow b feet long. Find the 
angle when a = 120, b = 70. 

66. How high is a tree that casts a horizontal shadow b feet 
in length when the angle of elevation of the sun is A ? Find 
the height of the tree when b = 80, A = 50°. 

U 67. What is the angle of elevation of an inclined plane if it 
rises 1 foot in a horizontal distance of 40 feet ? 

68. A ship is sailing due north-east with a velocity of 10 
miles an hour. Find the rate at which she is moving due 
north, and also due east. 

69. In front of a window 20 feet high is a flower-bed 6 feet 
wide. How long must a ladder be to reach from the edge of 
the bed to the window ? 

70. A ladder 40 feet long may be so placed that it will reach 
a window 33 feet high on one side of the street, and by turn- 
ing it over without moving its foot it will reach a window 21 
feet high on the other side. Find the breadth of the street. 

71. From the top of a hill the angles of depression of two 
successive milestones, on a straight level road leading to the 
hill, are observed to be 5° and 15°. Find the height of the 
hill. 

72. A fort stands on a horizontal plain. The angle of ele- 
vation at a certain point on the plain is 30°, and at a point 100 
feet nearer the fort it is 45°. How high is the fort ? 

73. From a certain point on the ground the angles of eleva- 
tion of the belfry of a church and of the top of the steeple were 
found to be 40° and 51° respectively. From a point 300 feet 
farther off, on a horizontal line, the angle of elevation of the 
top of the steeple is found to be 33° 45'. Find the distance 
from the belfry to the top of the steeple. 

74. The angle of elevation of the top of an inaccessible fort 
C, observed from a point A, is 12°. At a point B, 219 feet 
from A and on a line AB perpendicular to AC, the angle ABC 
is 61° 45'. Find the height of the fort. 



24 



TRIGONOMETRY. 



§ 17. The Isosceles Triangle. 

An isosceles triangle is divided by the perpendicular from 
the vertex to the base into two equal right triangles. 

Therefore, an isosceles triangle is determined by any two 
parts that determine one of these right triangles. 

Let the parts of an isosceles triangle ABC (Fig. 13), among 
which the altitude CD is to be in- 
cluded, be denoted as follows : 



a = one of the equal sides. 
c = the base. 
h = the altitude. 
A = one of the equal angles. 
C— the angle at the vertex. 






rig. 13. 


\c B 


For example : 
quired A, C, h. 


Given a 


and c 


1. 


cos A = 


\c c 

a 2 a 








2. 


C+2A 


= 180°; . 


\c=mr-2A 


= 2(90°- 


-A). 



3. h may be found directly in terms of a and c from the 



equation 



A2 +x 

4 



which gives h = V (a — * c) (a + £ c). 

But it is better to find the angles first, and then find h from 
either one of the two equations, 

- = sin A, or — = tan A. 
a \c 

whence h = asmA, or h =%c\asaA. 



The numerical values of A, C, and h may be computed by 
the aid of logarithms, as in the case of the right triangle. 

The area F of the triangle may be found, when c and h are 
given or have been found, by means of the formula 
F=ick 



THE ISOSCELES TRIANGLE. 



25 



Exercise VII. 

In an isosceles triangle : 

1. Given a and A\ find C, c, h. 

2. Given a and O; find A, c, h. 

3. Given c and A ; find C, a, h. 

4. Given c and O; find A, a, h. 

5. Given h and A ; find C, a, c. 

6. Given h and (7; find J., a, c. 

7. Given a and h ; find J., C, c. 

8. Given <? and h ; find ^4, C, a. 

9. Given a = 14.3, c = 11 ; find J., C, A. 

10. Given a = 0.295, J. = 68° 10' ; find c, h, F 

11. Given c = 2.352, C=69°49'; find a, A, ^ 

12. Given h = 7.4847, A = 76° 14' ; find a, <?, .# 

13. Given a = 6.71, A = 6.60; find A, C, c. 

14. Given c = 9, A = 20 ; find .4, (7, a. 

15. Given c = 147, ,P= 2572.5 ; find A, C, a, h. ■ 

16. Given h = 16.8, F= 43.68 ; find A, C, a, c. 

17. Find the value of i^in terms of a and c. 

18. Find the value of i^in terms of a and C. 

19. Find the value of Fin terms of a and A. 

20. Find the value of F in terms of h and O. 

21. A barn is 40 X 80 feet, the pitch of the roof is 45° ; find 
the length of the rafters and the area of both sides of the roof. 

22. In a unit circle what is the length of the chord corre- 
sponding to the angle 45° at the centre ? 

23. If the radius of a circle = 30, and the length of a chord 
«= 44, find the angle at the centre. 

24. Find the radius of a circle if a chord whose length is 5 
subtends at the centre an angle of 133°. 

25. What is the angle at the centre of a circle if the corre- 
sponding chord is equal to f of the radius ? 

26. Find the area of a circular sector if the radius of th^ 
circle = 12, and the angle of the sector = 30°, 



26 



TRIGONOMETRY. 



§ 18. The Regular Polygon. 

Lines drawn from the centre of a regular polygon (Fig. 14) 
to the vertices are radii of the circumscribed circle ; and lines 
drawn from the centre to the middle points of the sides are 
radii of the inscribed circle. These lines divide the polygon 
into equal right triangles. Therefore, a regular polygon is 
determined by a right triangle whose sides are the radius of 
the circumscribed circle, the radius of the inscribed circle, and 
half of one side of the polygon. 

If the polygon has n sides, the angle of this right triangle at 
the centre is equal to 

1 /360°\ 180° 

- f or 

2\ n J n 

If, also, a side of the polygon, or one of the above-mentioned 
radii, is given, this triangle may be solved, and the solution 
gives the unknown parts of the polygon. 

Let, 
n = number of sides. 
c == length of one side. 
r = radius of circumscribed circle. 
h = radius of inscribed circle. 
p = the perimeter. 
F= the area. 

Then, by Geometry, 

F= £ tip. 

Fig. 14. 








Exercise VIII. 


1. 


Given n = 10, 


c = l; 


find r, h, F. 


2. 


Given n = 12, 


p=70; 


find r, h, F. 


3. 


Given n = 18, 


r = l; 


find h,p, F. 


4. 


Given n = 20, 


r = 20, 


find h, c, F 


5. 


Given n = 8, 


h = l; 


find r, c, F 


6. 


Given w = ll, 


F=20; 


find r, A, c. 


7. 


Given w = 7, 


F=7; 


find r, h, p. 



THE REGULAR POLYGON. - 27 

8. Find the side of a regular decagon inscribed in a unit 
circle. 

9. Find the side of a regular decagon circumscribed about, 
a unit circle. 

10. If the side of an inscribed regular hexagon is equal to 1, 
find the side of an inscribed regular dodecagon. 

11. Given n and c, and let b denote the side of the inscribed 
regular polygon having 2n sides; find b in terms of n and c. 

12. Compute the difference between the areas of a regular 
octagon and a regular nonagon if the perimeter of each is 16. 

13. Compute the difference between the perimeters of a 
regular pentagon and a regular hexagon if the area of each is 12. 

14. From a square whose side is equal to 1 the corners are 
cut away so that a regular octagon is left. Find the area of 
this octagon. 

15. Find the area of a regular pentagon if its diagonals are 
each equal to 12. 

16. The area of an inscribed regular pentagon is 331.8; 
find the area of a regular polygon of 11 sides inscribed in the 
same circle. 

17. The perimeter of an equilateral triangle is 20; find 
the area of the inscribed circle. 

18. The area of a regular polygon of 16 sides, inscribed in 
a circle, is 100 ; find the area of a regular polygon of 15 sides, 
inscribed in the same circle. 

19. A regular dodecagon is circumscribed about a circle, 
the circumference of which is equal to 1 ; find the perimeter 
of the dodecagon. 

20. The area of a regular polygon of 25 sides is equal tc 
40 ; find the area of the ring comprised between the circum- 
ferences of the inscribed and the circumscribed circles. 



CHAPTEE III. 



GONIOMETRY. 

§ 21. Definition of Goniometry. 

In order to prepare the way for the solution of an oblique 
triangle, we now proceed to extend the definitions of the 
trigonometric functions to angles of all magnitudes, and to 
deduce certain useful relations of the functions of different 
angles/ 

That branch of Trigonometry which treats of trigonometric 
functions in general, and of their relations, is called Goniometry. 



§ 22. Angles of any Magnitude. 

Let the radius OP of a circle (Fig. 16) generate an angle by 

turning about the centre 0. This 
angle will be measured by the 
arc described by the point P; 
and it may have any magnitude, 
because the arc described by P 
may have any magnitude. 

Let the horizontal line OA be 
the initial position of OP, and 
let OP revolve in the direction 
shown by the arrow, or opposite 
to the way clock-hands revolve. 
Let, also, the four quadrants into 
which the circle is divided by the horizontal and vertical 
diameters AA', BB\ be numbered I., II., III., IV., in the 
direction of the motion. 




GONIOMETRY. „ 29 



During one revolution OP will form with OA all angles from 
0° to 360°. Any particular angle is said to be an angle of the 
quadrant in which OP lies ; so that, 

Angles between 0° and 90° are angles of Quadrant I. 

Angles between 90° and 180° are angles of Quadrant II. 

Angles between 180° and 270° are angles of Quadrant III. 

Angles between 270° and 360° are angles of Quadrant IV. 

If OP make another revolution, it will describe all angles 
from 360° to 720°, and so on. 

If OP, instead of making another revolution in the direction 
of the arrow, be supposed to revolve backwards about O, this 
backward motion tends to undo or cancel the original forward 
motion. Hence, the angle thus generated must be regarded 
as a negative angle ; and this negative angle may obviously 
have any magnitude. Thus we arrive at the conception of an 
angle of any magnitude, positive or negative. 

§ 23. General Definitions of the Functions. 

The definitions of the trigonometric functions may be ex- 
tended to all angles, by making the functions of any angle 
equal to the line values in a unit circle drawn for the angle 
in question, as explained in § 3. But the lines that represent 
the sine, cosine, tangent, and cotangent must be regarded as 
negative, if they are opposite in direction to the lines that repre- 
sent the corresponding functions of an angle in the first quad- 
rant ; and the lines that represent the secant and cosecant must 
be regarded as negative, if they are opposite in direction to the 
moving radius. 

Figs. 17-20 show the functions drawn for an angle A OP in 
each quadrant taken in order. In constructing them, it must 
be remembered that the tangents to the circle are always 
drawn through A and B, never through A f or B'. 

Let the angle AOPhe denoted by x; then, in each figure 
the absolute values of the functions, that is, their values with' 
out regard to the signs + and — , are as follows : 



30 



TRIGONOMETRY. 



sin x = MP, 

cos# = OM, 

B 



tan#:= A T, 

cot x = US, 

AN 




Keeping in mind the position of the points A and J3, we may 
define in words the first four functions of the angle x thus : 
the vertical projection of the moving radius ; 
the horizontal projection of the moving radius ; 
the distance measured along a tangent to the circle 
from the beginning of the first quadrant to the 
moving radius produced ; 
the distance measured along a tangent to the circle 
from the end of the first quadrant to the moving 
radius produced, 
and csc:r are the distances from the centre of the 
circle measured along the moving radius produced to the tan- 
a, gent and cotangent respectively. 



sin x = 
cos x = 

tan# = 

cot# = 

Sec a: 



GONIOMETRY. 



31 



§ 24. Algebraic Signs of the Functions. 

The lengths of the lines, defined above as the functions of 
any angle, are expressed numerically in terms of the radius 
of the circle as the unit. But, before these lengths can be 
treated as algebraic quantities, they must have the sign -f or 
— prefixed, according to the condition stated in § 23. 

The reason for this condition lies in that fundamental rela- 
tion between algebraic and geometric magnitudes, in virtue of 
which contrary signs in Algebra correspond to opposite direc- 
tions in Geometry. 

The sine MP and the tangent A T always extend from the 
horizontal diameter, but sometimes upwards and sometimes 
downwards ; the cosine OM and the cotangent B8 always 
extend from the vertical diameter, but sometimes towards the 
right and sometimes towards the left. The functions of an angle 
in the first quadrant are assumed to be positive. Therefore, 

1. Sines and tangents extending from the horizontal diam- 
eter upwards, are positive ; downwards, negative. 

2. Cosines and cotangents extending from the vertical diame- 
ter towards the right, are positive ; towards the left, are nega- 
tive. 

The signs of the secant and cosecant are always made to 
agree with those of the cosine and sine respectively. This 
agreement is secured if secants and cosecants extending from 
the centre, in the direction of the moving radius, are consid- 
ered positive ; in the opposite direction, negative. 

Hence, the signs of the functions for each quadrant are : 



Sine and cosecant 


I. 


II. 


III. 


IV. 


+ 


+ 


— 


— 


Cosine and secant 


+ 


— 


— 


+ 


Tangent and cotangent .... 


+ 


— 


+ 


— 



32 



7?.:>;::;vzi?.y 



§25. Functions of a Vabiable Angle. 

Let the angle x increase continuously from 0° to 360°; 
what changes will the valnes of its functions undergo ? 

It is easy, by reference to Figs. 21-24, to trace these 
changes throughout all the quadrants. 







1. Ttie Sine. In the first quadrant, the sine J/Pinc: 
from to 1 ; in the second, it remains positive, and decreases 
from 1 to ; in the third, it is negative, and increases in abso- 
lute value from to 1 ; in the fourth, it is negative, and 
decreases in absolute value from 1 to 0. 



GONIOMETRY. 33 



2. The Cosine. In the first quadrant, the cosine OM de- 
creases from 1 to 0; in the second, it becomes negative and 
increases in absolute value from to 1 ; in the third, it is 
negative and decreases in absolute value from 1 to 0; in the 
fourth, it is positive and increases from to 1. 

3. The Tangent. In the first quadrant, the tangent AT 
increases from to co; in the second quadrant, as soon as 
the angle exceeds 90° by the smallest conceivable amount, the 
moving radius OP, prolonged in the direction opposite to that 
of OP, will cut A T at a point T situated very far below A ; 
hence, the tangents of angles near 90° in the second quad- 
rant have very large negative values. As the angle increases, 
the tangent discontinues negative but diminishes in absolute 
value. When # = 180°, then T coincides with A, and tan 180° 
= 0. In the third quadrant, the tangent is positive and in- 
creases from to co ; in the fourth, it is negative and decreases 
in absolute value from co to 0. 

4. The Cotangent. In the first quadrant, the cotangent BS 
decreases from co to ; in the second quadrant, it is negative 
and increases in absolute value from to co ; in the third and 
fourth quadrants, it has the same sign, and undergoes the same 
changes as in the first and second quadrants respectively. 

5. The Secant. In the first quadrant, the secant O Tin- 
creases from 1 to co ; in the second quadrant, it becomes 
negative (being measured in the direction opposite to that of 
OP), and decreases in absolute value from co to 1, so that 
sec 180° = — 1 ; in the third quadrant, it continues negative, 
and increases in absolute value from 1 to co ; in the fourth 
quadrant, it is positive, and decreases from oo to 1. 

6. The Cosecant. In the first quadrant, the cosecant OS 
decreases from co to 1 ; in the second quadrant, it remains 
positive, and increases from 1 to co ; in the third quadrant, it 
becomes negative, and decreases in absolute value from co to 
1, so that esc 270° = — 1; in the fourth quadrant, it is nega- 
tive, and increases in absolute value from 1 to co. 



34 



TRIGONOMETRY. 



The limiting values of the functions are as follows : 



Sine 

Cosine 

Tangent 

Cotangent 

Secant . 

Cosecant 


0° 


90° 


180° 


270° 


360* 


±0 
1 

db 

rbOO 

1 
± 00 


1 

±0 

± 00 

±0 

± 00 

1 


±0 

-1 

± 

zb 00 

-1 

rb 00 


-1 
±0 

± 00 

±0 

dh 00 

-1 


±0 

1 

±0 

rb 00 

1 
zb 00 



Sines and cosines extend from + 1 to — 1 ; tangents and co- 
tangents from +oo to — oo ; secants and cosecants from -f oo 
to + 1) and from — 1 to — oo. 

In the table given above the double sign ± is placed before and 
oo. From the preceding investigation it appears that the functions always 
change sign in passing through and oo ; and the sign + or — prefixed 
to or oo simply shows the direction from which the value is reached. 

Take, for example, tan 90° : The nearer an acute angle is to 90°, the 
greater the positive value of its tangent; and the nearer an obtuse angle 
is to 90°, the greater the negative value of its tangent. When the angle 
is 90°, OP (Fig. 21) is parallel to AT, and cannot meet it. But tan 90° 
may be regarded as extending either in the positive or in the negative 
direction ; and according to the view taken, it will be + oo or — oo. 



§ 26. Functions of Angles Larger than 360°. 

It is obvious that the functions of 360° -f- x are the same 
both in sign and in absolute value as those of x ; for the mov- 
ing radius has the same position in both cases. In general, if 
n denote any positive whole number, 

The functions of (nX 360° -f- x) are the same as those of x. 

For example : the functions of 2200° = the functions of 
(6 X 360° + 40°) = the functions of 40°. 



GONIOMETRY. 35 



I 



§ 27. Extension of Formulas [l]-[3] to all Angles. 

The Formulas established for acute angles in § 5 hold true 
for all angles. Thus, Formula [1], 

sin 2 # + cos 2 :r = 1, 
is universally true ; for, whether MP and OM (Figs. 21-24) 
are positive or negative, MP** and OM 2 are always positive, 
and in each quadrant MP 2 + OM 2 = OP = 1. 

Also, Formulas 

r n sin x 

2 tan# = , 

L cosa? 

{sin xX csc# = l, 
cosrr X sec# = l, 
tan#X cot# = l, 
are universally true ; for the algebraic signs of the functions, 
as given in the table at the end of § 24, agree with those in 
Formulas [2] and [3] ; and with regard to the absolute values, 
we have in each quadrant from the similar triangles OMP, 
OAT, OPS, (Figs. 21-24) the proportions 

AT :G>A=MP:OM, 

MP : OP = OP :OS t 

OM:OP=OA :OT, 

AT :OA = OP :PS, 
which, by substituting 1 for the radius, and the right names 
for the other lines, are easily reduced to the above formulas. 
Formulas [l]-[3] enable us, from a given value of one func- 
tion, to find the absolute values of the other five functions, and 
also the sign of the reciprocal function. But in order to deter- 
mine the proper signs to be placed before the other four 
functions, we must know the quadrant to which the angle in 
question belongs ; or what amounts to the same thing, the sign 
of any one of these four functions ; for, by reference to the 
Table of Signs (§ 24) it will be seen that the signs of any two 
functions that are not reciprocals determine the quadrant to 
which the angle belongs. 



36 TRIGONOMETRY. 



Example. Given sin# = -f £, and tan x negative ; find the 
values of the other functions. 

Since sin a; is positive, x must be an angle in Quadrant I. or 
in Quadrant II.; but, since also tana; is negative, Quadrant I. 
is inadmissable. 

By [1], cosa; = ±Vl--Jf = ± f. 

Since the angle is in Quadrant II. the minus sign must be 
taken, and we have 

cosa; = — |. 
By [2] and [3], 

tano: = — ^|, Q,otx= — \, sec:r = — -J, csc# = J. 



Exercise IX. 

1. Construct the functions of an angle in Quadrant II. 
What are their signs ? 

2. Construct the functions of an angle in Quadrant III. 
What are their signs ? 

3. Construct the functions of an angle in Quadrant IV. 
What are their signs ? 

4. What are the signs of the functions of the following 
angles: 340°, 239°, 145°, 400°, 700°, 1200°, 3800°? 

5. How many angles less than 360° have the value of the 
sine equal to -f-y, and in what quadrants do they lie? 

6. How many values less than 720° can the angle x have 
if cos a; = -f- •§-, and in what quadrants do they lie ? 

7. If we take into account only angles less than 180°, how 
many values can x have if sin# = -f- ? if cos# = \ ? if cos a; = 
-f? if tanrr = |? if cota; = -7? 

8. Within what limits must the angle x lie if cos# = — ■§•? 
if cot x = 4 ? if sec x = 80 ? if esc x = — 3 ? (x to be less than 
360°.) 

9. In what quadrant does an angle lie if sine and cosine 
are both negative ? if cosine and tangent are both negative ? 
if the cotangent is positive and the sine negative ? 



GONIOMETRY. 37 



10. Between 0° and 3600° how many angles are there whose 
sines have the absolute value f ? Of these sines how many- 
are positive and how many negative ? 

11. In finding cos a? by means of the equation cos# = 
±Vl — sm 2 x, when must we choose the positive sign and when 
the negative sign ? 

12. Given cos x = — y/\ ; find the other functions when x is 
an angle in Quadrant II. 

13. Given tan# = V3 ; find the other functions when x is 
an angle in Quadrant III. 

14. Given sec# = -f 7, and tanrr negative; find the other 
functions of x. 

15. Given cot# = — 3; find all the possible values of the 
other functions. 

16. What functions of an angle of a triangle may be nega- 
tive ? In what case are they negative ? 

17. What functions of an angle of a triangle determine the 
angle, and what functions fail to do so ? 

18. Why may cot 360° be considered equal either to + co 
or to — oo ? 

19. Obtain by means of Formulas [l]-[3] the other func- 
tions of the angles given : 

(i.) tan 90° = oo. (iii.) cot 270° = 0. 

(ii.) cos 180° = -1. (iv.) csc3G0° = -oo. 

20. Find the values of sin 450°, tan 540°, cos 630°, cot 720°, 
sin 810°, esc 900°. 

21. For what angle in each quadrant are the absolute values 
of the sine and cosine alike ? 

Compute the values of the following expressions : 

22. a sin 0° + b cos 90° -c tan 180°. 

23. a cos 90° -h tan 180° + c cot 90°. 

24. a sin 90° - b cos 360° + (a - b) cos 180°. 

25. (a 2 - b 2 ) cos 3G0° - 4 ab sin 270°. 



38 



TRIGONOMETRY. 




§ 28. Reduction of Functions to the First Quadrant. 

In a unit circle (Fig. 25) draw two diameters PP and QS 

equally inclined to the horizon- 
tal diameter A A', or so that the 
angles AOP, A'OQ, A' OP, and 
AOS shall be equal. From the 
points P, Q, P, S let fall per- 
pendiculars to AA' ; the four 
right triangles thus formed, with 
a common vertex at 0, are equal ; 
because they have equal hypote- 
nuses (radii of the circle) and 
equal acute angles at 0. There- 
fore, the perpendiculars PIT, 
QN, PJ\ T , SM, are equal. Now these four lines are the sines 
of the angles AOP, AOQ, AOP, and AOS, respectively. 
Therefore, in absolute value, 

sin AOP = sin AOQ = sin AOP = sin AOS. 

And from § 27 it follows that in absolute value the cosines 
of these angles are also equal ; and likewise the tangents, the 
cotangents, the secants, and the cosecants.* 

Hence, for every acute angle (AOP) there is an angle in each 
of the higher quadrants whose functions, in absolute value, are 
equal to those of this acute angle. 

Let ZAOP=x, ZPOB = y; then x + y^W, and the 
functions of x are equal to the co-named functions of y (§ 4) ; 
further, 

Z AOQ (in Quadrant II.) = 180° -x= 90° + y, 
Z AOP (in Quadrant III.) = 180° -f x = 270° - y, 
A AOS (in Quadrant IV.) = 360° - x = 270° + y. 
Hence, if we prefix in each case the proper sign (§ 24), we 
have the two following series of Formulas : 



* In future, secants and cosecants will be disregarded. They may be found 
by [3] if wanted, but are seldom used in computations. 



GONIOMETRY. 39 



Angle in Quadrant II 

sin (180° — x) = sin x. sin (90° + y) = cos y. 

cos (180° — x) = — cos a?. cos (90° + y) — — sin y. 

tan (180° — a;) = — tan a;, tan (90° -f- y) = — cot y. 

cot (180° - x) = - cot a;. cot (90° + y) = - tany. 

Angle in Quadrant III 

sin (180° -f a?) = — sin x. sin (270° — y) = — cos y. 

cos (180° -f- x) = — cos a\ cos (270° — y) = — sin y. 

tan (180° + x) = tana;. tan(270°-y)= coty. 

cot (180° + x) = cot a;. cot(270°-y)= tany. 

Angle in Quadrant IV. 

sin (360° — a?) == — sin a?. sin (270° + y) = - cos y. 

cos (360° — x) = cos a;. cos (270° -f y) = sin y. 

tan (360° - a?) = - tana;. tan (270° + y) = - cot y. 

cot (360° - a?) = - cot x. cot (270° + y) = - tany. 

Hence, by selecting the right formulas, 

The functions of all angles can be reduced to the functions of 
angles not greater than 45°. Thus, to find the functions of 
220° and 230°, we should consider 220° as (180° + 40°), but 
230° as (270° - 40°). 

It is evident from these formulas that, 

If an acute angle be added to or subtracted from 180° or 360° f 
the functions of the resulting angle are equal in absolute value 
to the like-named functions of the acute angle; but if an acute 
angle be added to or subtracted from 90° or 270°, the functions 
of the resulting angle are equal in absolute value to the co-named 
functions of the acute angle. 

It is evident from the formulas for (180° — x) that, 

A given value of a sine determines two supplementary angles, 
one acute, the other obtuse ; a given value of any other function 
{except the cosecant) determines only one angle : acute if the 
value is positive, obtuse if the value is negative. 



e 



40 



TRIGONOMETRY. 




§ 29. Angles whose Difference is 90°. 

The general form of two such angles is x and 90° -f x, and 
they must lie in adjoining quadrants. The relations between 

their functions were found in 
§ 28, but only for the case when 
x is acute. These relations, how- 
ever, may be shown to hold true 
for all values of x. 

In a unit circle (Fig. 26) draw 
two diameters PR, and QS per- 
pendicular, to each other, and 
let fall to A A' the perpendicu- 
lars P3f, QE, RK, and SK 
The right triangles OMP, OHQ, 
OKR, and ONS are equal, because they. have equal hypote- 
nuses and equal acute angles POM, OQH, ROK, and OSN. 

Therefore, OM= QS= OK = ITS, 

and PM= OJ?= KR = ON. 

Hence, taking also into account the algebraic sign, 

sin AOQ= cos AOP; sin A08 = cos AOR; 

cos AOQ = - sin AOP; cos AOS =- sin AOR: 

sin AOR = cos AOQ; sin (360° + A OP) = cos AOS; 

cos AOR = - sin AOQ ; cos (360° + AOP) == - sin AOS. 

In all these equations, if rr denote the angle on the right-hand 
side, the angle on the left-hand side will be 90° + x. There- 
fore, if x be an angle in any one of the four quadrants, 

sin (90° -\~x)= cos x, 

cos (90° + x) = — sin x. 

And, by § 27, tan (90° + x) = - cot x, 

cot (90° + x) = — tana;. 

In like manner, it can be shown that all the formulas of 
§ 28 hold true, whatever be the value of the angle x. 



GONIOMETRY. 41 



1 

§ 30. Functions of a Negative Angle. 

If the angle A OP (Fig. 25) is denoted by x, the equal angle 
AOS, generated by a backward rotation of the moving radius 
from the initial position OA, will be denoted by — x. It is 
obvious that the position OS of the moving radius for this 
angle is identical with its position for the angle 360° — x. 
Therefore, the functions of the angle — x are the same as those 
of the angle 360° - x ; or (§ 28), 

sin (— x) — — sin x, tan (— x) = — tan x, 

cos (— x) = cos x, cot (— x) = — cot #. 



Exercise X. 

1. Express sin 250° in terms of the functions of an acute 
angle greater than 45°, and also in terms of the functions of 
an acute angle less than 45°. 

Ans. 1. sin 250° = sin (180° + 70°) = -sin 70°. 
2. sin 250° = sin (270° - 20°) = - cos 20°. 

Express the following functions in terms of the functions of 
angles less than 45° : 

2. sin 172°. 8. sin 204°. 14. sin 163° 49'. 

3. cos 100°. 9. cos 359°. 15. cos 195° 33'. 

4. tanl25°. 10. tan300°. 16. tan 269° 15'. 

5. cot 91°. 11. cot 264°. 17. cot 139° 17'. 

6. sec 110°. 12. sec 244°. 18. sec 299° 45'. 

7. esc 157°. 13. esc 271°. 19. esc 92° 25'. 

Express all the functions of the following negative angles in 
terms of those of positive angles less than 45° : 

20. -75°. 22. -200°. 24. -52° 37'. 

21. -127°. 23. -345°. 25. -196° 54'. 
26. Find the functions of 120°. 

Hint. 120° = 180° - 60°, or, 120° = 90° + 30° ; then apply g 28. 



42 TRIGONOMETRY. 



Find the functions of the following angles : 

27. 135°. 29. 210°. 31. 240°. 33. -30°. 

28. 150°. 30. 225°. 32. 300°. 34. -225°. 

35. Given sina; = — VJ, and cos x negative ; find the other 
functions of x, and the value of x. 

36. Given cot# = — V3, and x in Quadrant II. ; find the 
other functions of x, and the value of x. 

37. Find the functions of 3540°. 

38. "What angles less than 360° have a sine equal to — £ ? 
a tangent equal to — V3? 

39. Which of the angles mentioned in Examples 27-34 have 
a cosine equal to — VT? a cotangent equal to —-V3? 

40. What values of x between 0° and 720° will satisfy the 
equation sin x == + -§- ? 

41. In each of the following cases find the other angle be- 
tween 0° and 360° for which the corresponding function (sign 
included) has the same value: sin 12°, cos 26°, tan 45°, cot 72°; 
sin 191°, cosl20°, tan 244°, cot 357°. 

42. Given tan 238° = 1.6; find sin 122°. 

43. Given cos 333° = 0.89; find tan 117°. 
Simplify the following expressions : 

44. a cos (90° -z) + b cos (90° + x). 

45. m cos (90° -x) sin (90° -x). 

46. (a - b) tan (90° -x) + (a + b) cot (90° + x). 

47. a? + b 2 -2abcos(180°-x). 

48. sin (90° -f- x) sin (180° -f x) + cos (90° -f x) cos (180° - x). 

49. cos(180°+a;)cos(270 o -y)-sin(180 o +^)sin(270 o -y). 

50. tan x + tan (-y)- tan (180° — y). 

51. For what values of x is the expression sin x -f- cos x 
positive, and for what values negative ? Represent the result 
by a drawing in which the sectors corresponding to the nega- 
tive values are shaded. 

52. Answer the question of last example for sin x— cos x. 

53. Find the functions of (x — 90°) in terms of the functions 
of x. 

54. Find the functions of (x — 180°) in terms of the functions 
of x. 



etf 



GONIOMETRY. 



43 



(^ 




sin x; hence, FF~ sin x X O-D = sin # cosy. 
cos#; hence, QQ = cos#X CD = cos#siny. 



§ 31. Functions of the Sum of Two Angles. 

In a unit circle (Fig. 27) let the angle AOB = x, the angle 
BOC= y ; then the angle A00= 
x + y. 

In order to express sin(#-|-y) 
and cos (x -f- y) in terms of the 
sines and cosines of x and y, draw 
CFX OA, OF _L OB, DEI. OA, 
FG ± OF; then CD = siny, OF 
= cosy, and the angle FCG = 
the angle GFO = x. Also, 
sin (# + y) = £F= Z>^7+ <?£. 

oz>~ 

Therefore, sin (x -f y) = sinx cos y + cos x sin y. [4] 

Again, cos (x + y) = 0^= OF-FG. 

= cos#; hence, OF = cosx X OF = cos x cosy. 

OF J 

T)C 

= sin x ; hence, FG = sin a; X CD = sin# siny. 

Therefore, cos(x + y) = cos x cos y — sin x sin y. [5] 

In this proof x and y, and also the sum x -f- y, are assumed 

to be acute angles. If the sum 
x-\~y of the acute angles a? and 
y is obtuse, as in Fig. 28, the 
proof remains, word for word, 
the same as above, the only dif- 
ference being that the sign of 
Oi^will be negative, as FG is 

now greater than OF. The above formulas, therefore, hold 

true for all acute angles x and y. 




44 TRIGONOMETRY. 



If these formulas hold true for any two acute angles x and 
y, they hold true when one of the angles is increased by 90°. 
Thus, if for x we write x' = 90° + x, then, by § 29, 

sm(V-fy) = sin (90° -f- x + y) = cos (# + ;?/), 
c<as(V -f- y) = cos (90° -f x -f y) = — sin (# + y)- 

Hence, by [5], sin(V-f y) = cos a; cosy — sin a; siny, 
by [4], cos(V-|- y) = — sin x cosy — cos# siny. 

Now, by §29, cosa;= sin (90°-f#)= sin a;', 
sin x = — cos (90° -f- x) = — cos a;'. 

Therefore, by putting sin a;' for cos#, and —cos x' for sin a;, 
in the right-hand members of the above equations, 

sin (x 1 -f y) = sin a;' cos y -f- cos a;' sin y , 
cos(V-f y) = cosa;' cosy — sin _x' sin y. 

Hence, it follows that Formulas [4] and [5] are universally 
true. For they have been proved true for any two acute 
angles, and also true when one of these angles is increased by 
90° ; hence they are true for each repeated increase of one or 
the other angle by 90°, and therefore true for the sum of any 
two angles whatever. 

By §27, 

, , v sin (x + y) sin x cos y + cos x sin y 

tan (x + y) = 7 — r^-{ = — ! — = =-£• 

cos(x-f-y) cosa; cosy — sin # siny 

If we divide each term of the numerator and denominator of 
the last fraction by cos x cosy, and again apply § 27, we obtain 

, N tanx-f-tany rc1 

tan(x + y) = , — _n__ J -. [6] 

v ■' 1 — tanxtany 

In like manner, by dividing each term of the numerator and 
denominator of the value of cot (a; + y) by sin x siny, Ave obtain - 

_£> \ rj] cotx + coty l J 



GONIOMETRY. 



45 



§ 32. Functions of the Difference of Two Angles. 

In a unit circle (Fig. 29) let the angle AOD = x, the angle 
COB = y ; then the angle AOC= 
x — y. 

In order to express sin(^ — y) 
and cos (x — y) in terms of the 
sines and cosines of x and y, draw 
CF±OA, CD±OD, DEI. OA, 
DG±FC prolonged ; then CD = 
siny, OD = cosy, and the angle 
DCG=ihe angle EDC=x. And, 
sin(a;-y)= CF= DE- CG. 
DE 
OD 
CG 
CD 
Therefore, sin (x — y) = sin x cos y — cos x sin y. [8] 

Again, cos (a; - y) = OF= OE+ DG. 

O T? 

- - - = cos a; ; hence, OE = cos a: X OD = cos x cosy. 

OD * 

DG 




= sin x ; 



= cos a; 



hence, DE= sin x X OD = sin x cosy. 

hence, CG = cos a? X CJ9 = cos x siny. 
sinx cosy — cosx siny. 



CD 



= sin x ; hence, DG = sin .r X CD 



sinx siny. 



Therefore , cos (x — y) = cos x cos y -f sin x sin y. 



[9] 



In this proof, both x and y are assumed to be acute angles ; 
but, whatever be the values of x and y, the same method of 
proof will always lead to Formulas [8] and [9], when due 
regard is paid to the algebraic signs. 

The general application of these formulas may be at once 
shown by deducing them from the general formulas established 
in § 31, as follows : 

It is obvious that (x — y)~hy~ x. If we apply Formulas 
[4] and [5] to (x - y)-\-y, then 



46 TRIGONOMETRY. 



sin {(x — y)-\-y\ or sin x = sin {x — y) cosy -f- cos (x — y) siny, 
eos{(a; — y) + yj or cosa; — cos (a; — y) cosy — sin {x — y) siny. 

Multiply the first equation by cosy, the second by siny, 

sin x cosy = sin (x — y) cos 2 y + cos (x — y) sin y cosy, 
cos x siny == — sin (# — y) sin 2 y -f- cos (a; — y) siny cosy ; 

whence, by subtraction, 

sin x cosy — cosa? sin y = sin (x — y) (sin 2 y -f- cos 2 y). 

But sin 2 y -f- cos 2 y = 1 ; therefore, by transposing, 

sin (x — y) = sin a; cosy — cosa; siny. 

Again, if we multiply the first equation by sin y, the second 
equation by cosy, and add the results, we obtain, by reducing, 

cos (x — y) = cos x cos y -f- sin x sin y. 

Therefore, Formulas [8] and [9], like [4] and [5], from which 
they have been derived, are universally true. 

From [8] and [9], by proceeding as in § 31, we obtain 

tanx-tany_ t [1Q] 

K n 1 + tanxtany L J 

. , N cotxcoty-f-1 m-, 

cot(x-y) = — ^-. [11] 

* J coty — cotx 

Formulas [4] -[11] may be combined as follows: 
sin (x ± y) = sin x cosy ± cos x siny, 
cos (x±y) = cos a; cosy =f sin a; siny, 



, . tana; ± tan y 

tan (x ± y) = = — — — > 

1^ tana; tan y 

., v cota; coty =F 1 

cot (x ±y) = r ^—7 — 

coty±cota; 



GONIOMETRY. 47 



§33. Functions of Twice an Angle. 

If, in Formulas [4] -[7], y- x, they become : 

sin 2x = 2sinxcosx. [12] cos2x = cos 2 x — sin 2 x. [13] 

n 2tanx ,-,.-, , n cot 2 x — 1 r , K -, 

tan2x = - — — y-i [14] cot2x = — — - — [15] 

1 — tan 2 x L J 2cotx 

By these formulas the functions of twice an angle are found 
when the functions of the angle are given. 

§ 34. Functions of Half an Angle. 

Take the formulas 

cos 2 :r + sin 2 x = 1 [1] 

cos 2 # — sin 2 # = cos 2 x [13] 



Subtract, 2sin 2 .r = 1 — cos 2x 

Add, 2 cos 2 # = 1 + cos 2 x 

Whence 



sm# = 



1 — cos 2 x |1 + cos2o: 



J- 



12 ~ \ 2 

These values, if z is put for 2x, and hence \z for x, become 

S inJ Z = ±J^^. [16] oo S H = ±spf^: [17] 

Hence, by division (§ 27), 

taniz = ±J5E^. [is] oofc*i=-a=J|±^l [19] 

\l + cosz J \1 — cosz L 

By these formulas the functions of half an angle may be 
computed when the cosine of the entire angle is given. 

The proper sign to be placed before the root in each case 
depends on the quadrant in which the angle \z lies. (§ 24.) 

Let the student show from Formula [18] that 

tan*J9=\|— -• (See page 19, Note 2.) 



48 TRIGONOMETRY. 



§ 35. Sums and Differences of Functions. 

From [4], [5], [8], and [9], by addition and subtraction : 

sin (x-\-y)-\- sin (x — y)— 2 sin a: cos y, 

sin (x + y) — sin (x — y) = 2 cos x sin y, 

cos (x -f- y) + cos (x — y)= 2 cos ar cos y, 

cos (a; + y) — cos (a: — y) = — 2 sin a; sin y ; 

or, by making x + y = A, and x — y = B, 

and therefore, a; = ^ ( A -f- J9), and 3/ = % (A — B), 

sinA + sinB= 2sin|(A + B)cos|(A-B). [20] 
sinA-sinB = 2cosi(A + B)sini(A-B). [21] 
cosA + cosB = 2cosi(A -f B)cos |(A - B). [22] 
cosA - cosB = - 2sini(A + B)sin i(A - B). [23] 
From [20] and [21], by division, we obtain 
sin A -f sin B 



sin A — sin B 
or, since cot ^(A — B) 



t&ni(A + B)coti(A - B) 



sin A + sin B _ tan J (A -f B) 
sin A — sin B tani(A — B)' 

Exercise XI. 



[24] 



1. Find the value of sin (x + y) and cos (a; -J- y), when sin a- 
= f , cos a; = f , siny = ^, cosy = if. 

2. Find sin (90° -y) and cos (90° — y) by making x =90° 
in Formulas [8] and [9]. 

Find, by Formulas [4]-[ll], the first four functions of: 

3. 90° + y. 8. 360° -y. 13. -y. 

4. 180° -y. 9. 360° + y. 14. 45° -y. 

5. 180° + y. 10. x -90°. 15. 45° + y. 

6. 270° -y. 11. x -180°. 16. 30° + y. 

7. 270° + y. 12. x -270°. 17. 60° -y. 



GONIOMETRY. 49 



18. Find sin 3 a? in terms of sina?. 

19. Find cos 3 a? in terms of cos x. 

20. Given tan -J a; = 1 ; "find cos x. 

21. Given cot \ x = V3 ; find sina?. 

22. Given sin a; = 0.2 ; find sin -J- a? and cos Jar. 

23. Given cos a; = 0.5 ; find cos 2 a; and tan 2 a;. 

24. Given tan 45° = 1 ; find the functions of 22° 30'. 

25. Given sin 30° = 0.5 ; find the functions of 15°. 

26. Prove that tan 18° = ^ 33° + sin 3° . 

cos 33 + cos 3 

Prove the following formulas : 

27. sin2a?= 2tan ^ ■ 29. tan±a? = 



l-ftan 2 a? 1 + cosa? 

28. cos2a;= 1 ~ ta A 30. coUa? = - Bmx ■ 

l + tan 2 a? 1 — cosa: 

31. sin|-a? ± eos^-a? = Vl'± sina?. 

QO tan x ± tan y 

oA. — . -£ = ± tana: tany. 

cot x ± cot y 

33. tan(45 e -a-) = 1 ~ tana; - 

v y 1 -f tana? 

If -4, i?, Care the angles of a triangle, prove that: 

34. sin A-\-sm i? + s m C~ 4 cos \ A cos|-i? cos^C. 

35. cos ^4 -J- cos i? + cos C= 1 -j- 4 sin -J- ^4 s in -|- i? sin % C. 

36. tan ^4 -f tani? + tan (7= tan A X tan I? X tan C. 

37. cot|-^. + cot|^ + cot|C / =coti^4xcot-i-^Xcot|a 

Change to forms more convenient for logarithmic computa- 
tion : 

38. cotar + tana?. 43. 1-f tana? tan y. 

39. cot a? — tan a?. 44. 1 — tan x tan y. 

40. cota? + tany. 45. cota?coty+L 

41. cot x — tan y. 46. cot a- cot y — 1. 
,p 1 — cos 2 a? »>r, tana? -f tan y 

1 + cos 2 a? cot a? -f- cot y 



CHAPTER IV. 

THE OBLIQUE TRIANGLE. 

§ 36. Law of Sines. 

Let A, B, C denote the angles of a triangle ABC (Figs. 30 
and 31), and a, b, c, respectively, the lengths of the opposite 
sides. 

Draw CB±AB, and meeting AB (Fig. 30) or AB pro- 
duced (Fig. 31) at D. Let CD = h. 




In both figures, — = sin A 
b 



In Fig. 30, 
In Fig. 31, 






h . D * 

- = sin B. 
a 



sin (180° — 5) = sin B. 



Therefore, whether h lies within or without the triangle, 
we obtain, by division, 

a _ sin A 
b sinB 



[25] 



THE OBLIQUE TRIANGLE. 



51 



By drawing perpendiculars from the vertices A and B to 
the opposite sides we may obtain, in the same way, 



sini? 

sin C 



a siiiwA 

c 



sin 



Hence the Law of Sines, which may be thus stated : 
The sides of a triangle are proportional to the sines of the 
opposite angles. 

If we regard these three equations as proportions, and take 
them by alternation, it will be evident that they may be writ- 
ten in the symmetrical form, 



sin A sin B sin O 



Each of these equal ratios has a simple geometrical mean- 
ing which will appear if the Law of Sines is proved as follows : 

Circumscribe a circle about the triangle ABC (Fig. 32), 
and draw the radii OA, OB, 00; 
these radii divide the triangle into 
three isosceles triangles. Let R 
denote the radius. Draw OM 
A- BO. By Geometry, the angle 
BOO ~2A) hence, the angle 
BOM=A, then BM=Rsm BOM 
= Bum A. 

.-. BO or a = 2 B sin A. 

In like manner, b = 2 B sin B, 
and c = 2B sin 0. Whence we 
obtain 

a b 




Fig. 32. 



2B = 



sin A sinB sin C 



That is : TJie ratio of ar\y side of a triangle to the sine of the 
opposite angle is numerically equal to the diameter of the cir* 
cumscribed circle. 



52 TRIGONOMETRY. 



Ir 



§ 37. Law of Cosines. 

This law gives the value of one side of a triangle in terms 
of the other two sides and the angle included between them. 

In Figs. 30 and 31, a 2 = h 2 + BD 2 . 

In Fig. 30, BD =c-AD; 

in Fig. 31, BD =AD-c; 

in both cases, BD 2 = AU -ZcxAD+c 2 . 

Therefore, in all cases, a 2 = h 2 + ALP- + c 2 -2cx AD. 

Now, h 2 + AD 2 = b 2 , 

and AD = b cos A. 

Therefore, a 2 = b 2 -f c 2 - 2 be cos A. [26] 

In like manner, it may be proved that 

b 2 = a 2 + c 2 — 2 ac cos B, 
c 2 = a 2 + b 2 -2abcosC. 

The three formulas have precisely the same form, and the 
law may be stated as follows : 

The square of any side of a triangle is equal to the sum of 
the squares of the other two sides, diminished by twice the 
product of the sides and the cosine of the included angle. 

§ 38. Law of Tangents. 

By § 36, a : b = sin A : sin B ; 

whence, by the Theory of Proportion, 

a— b sin A — sin B 

a-\-b sin A -f- sin B 

But by [24], page 48, 

sin A — sin B tan j (A — B) 

sin A -f sin B tan £ ( A -f B) 

Therefore, 

a — b _ tan \ (A — B) r^j j 

a + b-tanKA + B) L J 



THE OBLIQUE TRIANGLE. 53 

By merely changing the letters, 

a + c tan£(^+<7)' b + c tan *(£+#) 

Hence the Law of Tangents : 

The difference of two sides of a triangle is to their sum as the 
tangent of half the difference of the opposite angles is to the tan- 
gent of half their sum. 

Note. If in [27] b > a, then B>A. The formula is still true, but to 
avoid negative quantities, the formula in this case should be written 
b — a _ tan £ (B — A) 
b + a ~ tan £ (B + A) 

Exercise XII. 

1. What do the formulas of § 36 become when one of the 
angles is a right angle ? 

2. Prove by means of the Law of Sines that the bisector of 
an angle of a triangle divides the opposite side into parts pn> 
portional to the adjacent sides. 

3. What does Formula [26] become when A = 90° ? when 
A = 0° ? when A = 180° ? What does the triangle become in 
each of these cases ? 

Note. The case where A = 90° explains why the theorem of \ 37 is 
sometimes termed the Generalized Theorem of Pythagoras. 

4. Prove (Figs. 30 and 31) that whether the angle B is 
acute or obtuse c = a cos B -\- b cos A. What are the two sym- 
metrical formulas obtained by changing the letters? What 
does the formula become when B = 90° ? 

5. From the three following equations (found in the last 
exercise) prove the theorem of § 37 : 

c = a cos B -f- b cos A, 
b = a cos C -f- c cos A, 
a = b cos C -f c cos B. 
Hikt. Multiply the first equation by c, the second by b, the third 
by a ; then from the first subtract the sum of the second and third. 



54 



:: : .-_: ' \i:zzir: 



6. In Formnla [27] what is the maximum value of i {A— B) ? 

off (A+2)i 

7. Find the form to which Formula [27] reduces, and 
describe the nature of the triangle, when 

(i.) C= 90° ; (ii.) A - B = 90°. and B=C 

§ 39. The Give^t Pae.ts of jls Oblique Teia^tg-le. 

The formulas established in §> 36—38 together ~ iih the 

equation Jl — 1 — ?= 180°, are sufficient for solving every 
: : an oblique triangle. The three parts that determine 
an oblique triangle may be : 
I. One side and two angles 
II. Two sides and the angle opposite to one of these sides ; 
III. Two sides and the included angle ; 
IT. The thr^r a les 

In .-" JL. B. C denote the angles, a, b, : the ndes 

opposite these angles respectively. 



! « Iash I. 

- en erne side a, and two angles A and B ; find the remain- 
ing parts C 7 h, and c. 

i c=mr—(A+B). 

.1 a sin B a 



: 

a 

: __ 

a m 



suiB 
sin A 
sinC 



an A 

a sin C 



sin ^4 
a 



sin J 



X sin C. 



sinJL sin J. 
ExAMrLE =^il. A = ±: : l: B=23?1V. 

The work may be arranged as follow 



a= 24.31 

A = 45° IS' 

B= 22° ir 

A+B= 67° 29' 

;=::- c 3i' 



loga = l.:: r_ : 
cologrsin i = 0.1-:- : 

: ;■"" : 



- 
log- sin B 



log 6 = 1.11103 
*= 12.913 






=138578 

= 0.14525 
9.96556 



log <? = 1.49959 
-31.593 



THE OBLIQUE TRIANGLE. 55 







Exercise XIII. 




1. 


Given a = 


:500, 


A = 10° 12', 


^ = 46° 36'; 




find C= 


= 123° 12' 


, 5 = 2051.48, 


c = 2362.61. 


2. 


Given a = 


:795, 


A = 79° 59', 


^ = 44° 41'; 




find C= 


= 55° 20', 


b = 567.688, 


c = 663.986. 


3. 


Given a = 


•804, 


A = 99° 55', 


.£ = 45 o r; 




find 0= 


:35°4', 


5 = 577.313, 


c = 468.933. 


4. 


Given a = 


:820, 


A = 12° 49', 


^ = 141° 59'; 




find C= 


-- 25° 12', 


b = 2276.63, 


c = 1573.89. 


5. 


Given c — 


-- 1005, 


A = 78° 19', 


J5 = 54°27'; 




find (7= 


= 47° 14', 


a = 1340.6, 


5 = 1113.8. 


6. 


Given 5 = 


= 13.57, 


.5 = 13° 57', 


0=51° 13'; 




find .4 = 


= 108° 50' 


, a = 53.276, 


c = 47.324. 


7. 


Given a = 


= 6412, 


A = 70° 55', 


C= 52° 9' ; 




find ,8 = 


= 56° 56', 


5 = 5685.9, 


c = 5357.5. 


8. 


Given 5 = 


= 999, 


A = 37° 58', 


C= 65° 2'; 




find.£ = 


-77°, 


a = 630.77, 


c = 929.48. 



9. In order to determine the distance of a hostile fort A 
from a place B, a line i?C and the angles ABC and BCA 
were measured, and found to be 322.55 yards, 60° 34', and 
56° 10', respectively. Find the distance AB. 

10. In making a survey by triangulation, the angles B and 
of a triangle ^(7 were found to be 50° 30' and 122° 9', 
respectively, and the length i?(7is known to be 9 miles. Find 
AB and AC. 

11. Two observers 5 miles apart on a plain, and facing 
each other, find that the angles of elevation of a balloon in 
the same vertical plane with themselves are 55° and 58°, 
respectively. Find the distance from the balloon to each 
observer, and also the height of the balloon above the plain. 

12. In a parallelogram given a diagonal d and the angles 
x and y which this diagonal makes with the sides. Find the 
sides. Compute the results when d= 11.237, x= 19° 1', and 
y = 42°54. 



56 TRIGONOMETRY. 



13. A lighthouse was observed from a ship to bear N. 34° E. ; 
after sailing due south 3 miles, it bore N. 23° E. Find the dis- 
tance from the lighthouse to the ship in both positions. 

Note. The phrase to bear N. 34° E. means that the line of sight to 
the lighthouse is in the north-east quarter of the horizon, and makes, 
with a line due north, an angle of 34°. 

14. In a trapezoid given the parallel sides a and b, and the 
angles x and y at the ends of one of the parallel sides. Find 
the non-parallel sides. Compute the results when a = 15, 
5 = 7, * = 70 o , y = 40°. 

Solve the following examples without using logarithms : 

15. Given b = 7.07107, A = 30°, C= 105° ; find a and c. 

16. Given c = 9.562, A = 45°, B = 60° ; find a and b. 

17. The base of a triangle is 600 feet, and the angles at the 
base are 30° and 120°. Find the other sides and the altitude. 

18. Two angles of a triangle are, the one 20°, the other 40°. 
Find the ratio of the opposite sides. 

19. The angles of a triangle are as 5 : 10 : 21, and the side 
opposite the smallest angle is equal to 3. Find the other 
sides. 

20. Given one side of a triangle equal to 27, the adjacent 
angles equal each to 30°. Find the radius of the circum- 
scribed circle. (See § 36, Kemark.) 



§41. Case II. 

Given two sides a and b, and the angle A opposite to the 
side a ; find the remaining parts B, C, c. 

This case, like the preceding case, is solved by means of 

the Law of Sines. 

. sin B b , 1 n • -n b sin A . 

Since -: = — , therefore sin B = > 

sin A a a 



C= 180° -(A + B). 

c sin C 

a sin A 1 



A j . c sin O ,i j. a sin O 

And since -=- therefore c = — -• 



THE OBLIQUE TRIANGLE. 



57 



When an angle is determined by its sine it admits of two 
values, which are supplements of each other (§ 28) ; hence, 
either value of B may be taken unless excluded by the con- 
ditions of the problem. 

If a>b, then by Geometry A > B, and B must be acute 
whatever be the value of A ; for a triangle can have only 
one obtuse angle. Hence, there is one, and only one, triangle 
that will satisfy the given conditions. 

If a = b, then by Geometry A = B ; both A and B must be 
acute, and the required triangle is isoseeles. 

If a < b, then by Geometry A < B, and A must be acute 
in order that the triangle 
may be possible. If A is 
acute, it is evident from 
Fig. 33, where Z BA C=A, 
AC=b, CB = CB'=a, 
that the two triangles A CB 
and ACB' will satisfy the 
given conditions, provided 
a is greater than the per- 
pendicular CB; that is, 
provided a is greater than b sin A (§ 10). The angles ABC 
and AB'C are supplementary (since Z ABC= Z BB'C) ; 
they are in fact the supplementary angles obtained from the 




B 



Fig. 33. 



formula 



sin B = 



6 sin A 



If, however, a = l sin A = CP (Fig. 33), then sini?=l, 

B = 90°, and the triangle required is a. right triangle. 

If a < b sin A, that is, < CB, then sin B > 1, and the tri- 
angle is impossible. 

These results, for convenience, may be thus stated : 

If a > b, or a = b, or if a = b sin A, One solution. 

If a < b, but > b sin A, and A < 90°, Two solutions. 

or if 



Ifa<ft and^4>90 c 



a < b sin A and A < 90°, 
No solution. 



58 



TRIGONOMETRY. 



The number of solutions can often be determined by inspec- 
tion. If there is any doubt, it may be removed by computing 
the value of 6 sin A. 

Or we may proceed to compute log sin B. If log sin B = 0, 
the triangle required is a right triangle. If log sin B > 0, the 
triangle is impossible. If log sin B<. 0, there is one solution 
when a > b; there are two solutions when a<b. 

When there are two solutions, let B\ C, c' denote the un- 
known parts of the second triangle ; then, 

B'^180°-B, C = 180° -(A+B') = B - A, 

. a sin C 
smA 



Examples. 

1. Given a = 16, b = 20, A = 106° ; find the remaining 
parts. 

In this case a<b, and A > 90° ; therefore the triangle is impossible. 

2. Given a = 36, b = 80, A = 30° ; find the remaining 
parts. 

Here we have b sin A = 80 X ? = 40 ; so that a < b sin A, and the 
triangle is impossible. 

3. Given a= 72630,6 = 117480, J=80°0'50"; find B, C, c. 



a = 72630 
b = 117480 
A = 80° 0' 50" 



4. Given 



colog a = 5.13888 

log b = 5.06996 

log sin A = 9.99337 



Here log sin B>0. 
.'. no solution. 



lag sin B = 0.20221 

13.2, 5 = 15.7, A = 57° 13' 15.3"; find^, O, c. 



a = 13.2 
6 = 15.7 
A = 57° 13' 15.3" 


colog a = 8.87943 

log b = 1.19590 

log sin J. = 9.92467 

log sin B = 0.00000 
B = 90° 
.-. O = 32° 46' 44.7" 


c = b cos A 

log b =1.19590 

log cos ^1 = 9.73352 


Here log sin 5 = 0, 
,\ a right triangle. 


log c = 0.92942 
c = 8.5 



THE OBLIQUE TRIANGLE. 



59 



5. Given a =767, 6 = 242, A = 


36°53'2"; find .5, C, c. 


a =767 


colog a = 7.11520 




log a = 2.88480 


b =-242 


log 6 = 2.38382 


. 


log sin (7 = 9.86970 


A = 36° 53' 2" 


log sin ^1 = 9.77830 
log sin £ = 9.27732 




colog sin ^1 = 0.22170 


Here a > b, 


log c = 2.97620 


and log sin J5 < 0. 


B = 10° 54' 58 


it 


c = 946.675 


.*. one solution. 


.• % C=132°12'0" 




6. Given a = 177.01, 6 = 216.45, 


A = 35° 36' 20"; find the 


other parts. 




a =177.01 


colog a = 7.75200 


log a = 2.24800 


2.24800 


o = 216.45 


log b = 2.33536 


cologsin^i = 0.23493 


0.23493 


A = 35° 36' 20" 


log sin ^1=9.76507 
log sin £ = 9.85243 


lof 


rsinC=9.99462 


9.23034 


Here a < 6, 


log c = 2.47755 


1.71327 


and log sin B < 0. 


B = 45° 23' 28" 


c = 300.29 or 51.674 


.'.two solutions. 


or 134° 36' 32" 
.-. O = 99° 0' 12" 








or 9° 47' 8" 









Exercise XIV. 



1. 


Determine the number of solutions in each of the fol- 


lowin 


g cases : 






(i.) a = 80, 6 = 


= 100, A = 30°. 




(ii.) a = 50, 6 = 


- 100, A = 30°. 




(iii.) a = 40, 5 = 


= 100, A = 30°. 




(iv.) a = 13.4, 6 = 


= 11.46, A = 77° 20'. 




(v.) a =70, 6 = 


: 75, A = 60°. 




(vi.) a =134.16, b = 


= 84.54, .5 = 52° 9' 11". 


2. 


Given a = 840, 


6 = 485, A = 21° 31'; 




find JB = 12° 13' 34", 


(7= 146° 15' 26", £=1272.18. 


3. 


Given a = 9.399, 


6 = 9.197, A = 120° 35'; 




find B = 5T 23' 40", 


<7=2°1'20\ ^ = 0.38525. 


4. 


Given a = 91.06, 


6 = 77.04, ,4 = 51° 9' 6" 




find .5 = 41° 13', 


C= 87° 37' 54", c = 116.82. 


5. 


Given a = 55.55, 


6 = 66.66, ^=77° 44' 40" 




find A = 54° 31' 13", 


C= 47° 44' 7", £-=50.481. 



60 TRIGONOMETRY. 



6. Given a = 309, 5 = 360, A = 21° 14' 25"; 

find ^ = 24° 57' 54", (7= 133°47'41", c = 615.67, 
B'= 155° 2' 6", 0'= 3° 43' 29", c'= 55.41. 

7. Given a = 8.716, 5 = 9.787, A = 38° 14' 12"; 

find B = 44° 1' 28", C= 97° 44' 20", £ = 13.954, 
B'= 135° 58' 32", C'= 5° 47' 16", c'= 1.4203. 

8. Given a = 4.4, 5 = 5.21, ' ^L = 57° 37' 17"; 

find ^ = 90°, (7= 32° 22' 43", c = 2.79. 

9. Given a = 34, 5 = 22, ^ = 30° 20'; 

find A = 51° 18' 27", C= 98° 21' 33", c = 43.098, 

^'=128° 41' 33", C'= 20° 58' 27", c'= 15.593. 

10. Given 5 = 19, c = 18, Q- 15° 49' ; 

find B = 16° 43' 13", A = 147° 27' 47", a = 35.519, 
^'=163° 16' 47", ,4=0° 54' 13", a'= 1.0408,. 

11. Given a = 75, 5 = 29, B = 16° 15' 36" ; find the differ- 
ence between the areas of the two corresponding triangles. 

12. Given in a parallelogram the side a, a diagonal d, and 
the angle A made by the two diagonals ; find the other diag- 
onal. 

Special case: a = 35, d=63, .4 = 21° 36' 30". 

§ 42. Case III. 

Given two sides a and b and the included angle C '; find the 
remaining parts A, B, and c. 

Solution I. The angles A and B may both be found by 
means of Formula [27], § 38, which may be written 

tan %(A - B) = 2Ln| X tan K-^ + -#)• 
a-\- o 

Since J (A -f B) = | (180° - C), the value of \ (A + B) is 
known ; so that this equation enables us to find the value of 
i(A-B). We then have 

$(A + B) + l{A-B) = A t 

and i (A + B) - I \{A - B) = B. 



THE OBLIQUE TRIANGLE. 



61 



After A and B are known, the side c may be found by the 
Law of Sines, which gives its value in two ways, as follows : 



sinC 



b sin C 



sin A sin B 

Solution II. The third side c may be found directly from 
the equation (§ 37) 

c= Va 2 + b 2 -2abcosO; 

and then, by the Law of Sines, the following equations for 
computing the values of the angles A and B are obtained ; 



sin A = aX 



sinB = bX 



sin C 



Solution III. If, in the triangle ABC (Fig. 34), BD is 
drawn perpendicular to the side 
AC, then 

BD BD 



tan A 



Now 
and 



tan A = 



AD 
BD 

DC = a cos C 
a sin C 



AC- DC 

a sin C (§10), 



b — a cos C 
By merely changing the letters, 

b sin 




tani? = 



C 



It is not necessary, however, to use both formulas. When 
one angle, as A, has been found, the other, B, may be found 
from the relation A + B+ C= 180°. 

When the angles are known, the third side is found by the 
Law of Sines, as in Solution I. 

Note. When all three unknown parts are required, Solution I. is the 
most convenient in practice. When only the third side c is desired, Solu- 
tion II. may be used to advantage, provided the values of a 2 and Z> 2 can 
be readily obtained without the aid of logarithms. But Solutions II. 
and III. are not adapted to logarithmic work. 



62 



TRIGONOMETRY. 



Examples. 
1. Given a = 748, 5 = 375, (7= 63° 35' 30"; find A, B, 



and c. 






a + b = 


1123 




a — b = 


= 373 




CA+B)= 


116° 


24' 30" 


K^+B) = 


58° 


12' 15" 


${A-B) = 


28° 


10' 52" 


A = 


= 86° 


23' 7" 


B= 


30° 


1'23" 



log(a -&)== 2.57171 

colog(a + b) ==6.94961 

log tan £ (A+£) = 0.20766 

log tan £(^i-£) = 9.72898 
K^--S) = 28° 10' 52" 



log b = 2.57403 

log sin (7=9.95214 

colog sin B = 0.30073 



;c = 2.82690 
c = 671.27 



Note. In the above Example we use the angle B in finding the side 
c, rather than the angle A, because A is near 90°, and therefore its sine 
should be avoided. 

2. Given a = 4, c = 6, B = 60° ; find the third side J. 
Here Solution II. may be used to advantage. We have 
6= Va 2 + c 2 - 2ac cos _g= Vl6 + 36 - 24 = V28 ; 



that is, 



2accos.g=Vl6 + 36-24 
log 28 = 1.44716, log V28 = 0.72358, V28 = 5.2915; 
6 = 5.2915. 



If 



2. 



3. 



4. 



5. 





Exercise XV. 




Given a = 


= 77.99, 


5 = 83.39, C 


= 72° 15'; [§38,N. 


find .4 = 


= 51° 15', 


B = 56° 30', 


c = 95.24. 


Given 5 = 


= 872.5, 


c = 632.7, 


^4 = 80°; 


find B = 


= 60° 45', 


(7= 39° 15', 


a = 984.83. 


Given a - 


= 17, 


5 = 12, 


C=59°17'; 


find A = 


= 77° 12' 53", 


£ = 43° 30' 7", 


c = 14.987. 


Given 5 - 


= V5, 


c = V3, 


^ = 35° 53' ; 


find .5 = 


= 93° 28' 36", 


(7= 50° 38' 24", 


a = 1.313. 


Given a = 


= 0.917, 


5 = 0.312, 


(7= 33° 7' 9"; 


find A = 


= 132° 18' 27", 


^ = 14° 34' 24", 


c = 0.67748. 


Given a = 


-13.715, 


c= 11.214, 


^ = 15°22'36"i 


find A = 


= 118°55'49", 


C= 45° 41' 35", 


5 = 4.1554. 


Given b = 


= 3000.9, 


(7 = 1587.2, 


A = 86° 4' 4"; 


find£ = 


= 65° 13' 51", 


(7= 28° 42' 5", 


a = 3297.2. 



THE OBLIQUE TRIANGLE. 63 

8. Given a = 4527, 5 = 3465, C=66°6'27"; 

find A = 68° 29' 15", .£ = 45° 24' 18", c = 4449. 

9. Given a = 55.14, 5 = 33.09, C= 30° 24'; 

find ^4 = 117° 24' 33", B = 32° 11' 27", c = 31.431. 
10. Given a = 47.99, 5 = 33.14, 0= 175° 19' 10"; 

find A = 2° 46' 8", ^ = 1°54'42", c = 81.066. 
11.' If two sides of a triangle are each equal to 6, and the 
included angle is 60°, find the third side. 

12. If two sides of a triangle are each equal to 6, and the 
included angle is 120°, find the third side. 

13. Apply Solution I. to the case in which a = b or the 
triangle is isosceles. 

14. If two sides of a triangle are 10 and 11, and the in- 
cluded angle is 50°, find the third side. 

15. If two sides of a triangle are 43.301 and 25, and 'che 
included angle is 30°, find the third side. 

16. In order to find the distance between two objects A 
and B separated by a swamp, a station Cwas chosen, and the 
distances CA = 3825 yards, CB = 3475.6 yards, together with 
the angle ACB = 62° 31', were measured. Find the distance 
from A to B. 

17. Two inaccessible objects A and B are each viewed 
from two stations C and D 562 yards apart. The angle A CB 
is 62° 12', BCD 41° 8', ABB 60° 49', and ADC 34° 51'; 
required the distance AB. 

18. Two trains start at the same time from the same station, 
and move along straight tracks that form an angle of 30°, one 
train at the rate of 30 miles an hour, the other at the rate of 
40 miles an hour. How far apart are the trains at the end 
of half an hour ? 

19. In a parallelogram given the two diagonals 5 and 6, 
and the angle that they form 49° 18'. Find the sides. 

20. In a triangle one angle = 139° 54', and the sides form- 
ing the angle have the ratio 5 : 9. Find the other two angles. 



64 TKIGONOMETRY. 



§ 43. Case IV. 

Given the three sides a, b, c; find the angles A, B, O. 
The angles may be found directly from the formulas estab- 
lished in § 37. Thus, from the formula 

a 2 = b 2 + (?-2bccosA 

we nave cos A = ! — 

2 be 

From this equation formulas adapted to logarithmic work 
are deduced as follows : 

For the sake of brevity, let a-\-b -f- c = 2s; then b + c — a 
= 2 (s — a), a — b-j-c = 2(s — b), and a -f- b — c = 2 (s — c). 

Then the value of 1 — cos A is 

_ b 2 +c 2 -a 2 ^ 2bc-b 2 -e 2 + a 2 _ a 2 -(b-c) 2 
2be 2bc 2bc 

__ (a + b~c)(a-b + c) _ 2(s-b)(s-c) . 
2bc be 

and the value of 1 -f- cos A is 



1 b 2 + c 2 -a 2 ^ 2bc + b 2 + c 2 -a 2 _ = (b + c) i 






_ (b + g + a) (b + g - a) _ 2 s (s - a) 
2bc be 

But from Formulas [16] and [17], § 34, it follows that 

1 — cos A = 2 sin 2 i A, and 1 -f cos ^4 = 2 cos 2 ? A 



.<. 2sm 2 $A- 2 ( S ~ b )( s ~ c \ and 2cos 2 $A-- 
be 


_ 2 s (s — a) 
be 


whence sin|A = xr , ' 

\ DC 


[28] 


and therefore tan J A = A K s ~ b K s ~ c ) , 

\ s(B -a) 


[29] 
[30] 



THE OBLIQUE TRIANGLE. 65 

By merely changing the letters, 

• i t. j(s — a) (s — c) . . _. \(s — a)(s~ b) 
sin i B = \ - — -, sin £ C= \ '-\ L 



cos J^ = ^^> cos I (7= ^ 



s (s — c) 



<2& 

. _ l(s — a) (s — c) \(s — a)(s — b) 

tan J B = <\p ( ,K ,. J , tan i C= \^ ( JK . y - 

X s (s — o) \ s (s — c) 

There is then a choice of three different formulas for finding 
the value of each angle. If half the angle is very near 0°, 
the formula for the cosine will not give a very accurate result, 
because the cosines of angles near 0° differ little in value ; and 
the same holds true of the formula for the sine when half 
the angle is very near 90°. Hence, in the first case the 
formula for the sine, in the second that for the cosine, should 
be used. 

But, in general, the formulas for the tangent are to be 
preferred. 

It is not necessary to compute by the formulas more than 
two angles ; for the third may then be found from the equation 

A + B + C=180°. 

There is this advantage, however, in computing all three 
angles, by the formulas, that we may then use the sum of the 
angles as a test of the accuracy of the results. 

In case it is desired to compute all the angles, the formulas 
for the tangent may be put in a more convenient form. 

The value of tan I A may be written 



(s-a)(s-b)(s-e) or 1 k t - a )(s-b )(, -c) 
\ s(s — a) 2 s — a\ s 

Hence, if we put 



4 



(B- a)(B-p)(B-o) _ r> j- 31 -| 



we have tan|A= . [32] 

s — a 



6G 



TRIGONOMETRY. 



In like manner, 

tan J B 



s-b' 



tan J (7= 



Examples. 

1. Given a = 3.41, 5 = 2.60, c = 1.58 ; find the angles. 

Using Formula [30], and the corresponding formula for tan % B, we, 
may arrange the work as follows : 



a- 


= 3.41 


b- 


= 2.60 


c 


= 1.58 


2s 


= 7.59 


s 


= 3 795 


-a 


= 385 


-b 


= 1.195 


- c 


= 2.215 



colog s = 9.42079 




colog s= 9.42079-10 


colog(s — a) = 0.41454 




log(s-a)= 9.58546-10 


log(s- b) = 0.07737 




colog («-&)= 9.92263-10 


log(s-c) = 0.34537 




log (s-c) = 0.34537 


2)0.25807 




2)19.27425 - 20 


log tan £^ = 0.12903 




log tan J B= 9:63713-10 


\A= 53° 23' 20" 


\B = 23° 26' 37" 


A = 106° 46' 


40" 


£ = 4G D 53'14" 



.\A + B = 153° 39' 54", and 0« 26° 20' 6". 

2. Solve Example 1 by finding all three angles by the use 
of Formulas [31] and [32]. 

Here the work may be compactly arranged as follows, if we find 
log tan £-4, etc., by subtracting log(s — a), etc., from logr instead of 
adding the cologarithm : 



a = 3.41 


log(s- a) = 9.58546 


log 


tan £ A = 


10.12903 


b = 2.60 


log(s- b) = 0.07737 


log 


tan£# = 


9.63713 


c = 1.58 


log(s-c) = 0.34537 
colog s = 9.42079 


log 


tan 1 O = 


9.36912 


2s =7.59 


53° 23' 20" 


s = 3.795 

-a = 0.385 


log r 2 = 9.42899 
logr =9.71450 




A = 


23° 26' 37" 

13° 10' 3" 


-5 = 1.195 


100° 46' 40" 


- c = 2.215 


of). 




B = 

a= 


46° 53' 14" 


2s = 7.590 (pre 


26° 20' 0" 



Proof, A + B+C = 180° 0' 0" 

Note. Even if no mistakes are made in the work the sum of the 
three angles found as above may differ very slightly from 180° in conse- 
quence of the fact that logarithmic computation is at best only a method 
of close approximation. When a difference of this kind exists it should 
be divided among the angles according to the probable amount of error 
for each angle. 



THE OBLIQUE TRIANGLE. 



67 



Exercise XVI. 

Solve the following triangles, taking the three sides as the 
given parts : 



1 


a 


b 


c 


A 


B 





51 


65 


20 


38° 52' 48" 


126° 52' 12" 


14° 15' 


2 


78 


101 


29 


32° 10' 54" 


136° 23' 50" 


11° 25' 16" 


3 


111 


145 


40 


27° 20' 32" 


143° 7' 48" 


9° 31' 40" 


4 


21 


26 


31 


42° 6' 13" 


56° 6' 36" 


81° 47' 11" 


5 


19 


34 


49 


16° 25' 36" 


30° 24' 


133° 10' 24" 


6 


. 43 


50 


57 


46° 49' 35" 


57° 59' 44" 


75° 10' 41" 


7 


37 


58 


79 


26° 0'29" 


43° 25' 20" 


110° 34' 11" 


8 


73 


82 


91 


49° 34' 58" 


58° 46' 58" 


71° 38' 4" 


9 


14.493 


55.4363 


66.9129 


8° 20' 


33° 40' 


138° 


10 


V5 


V6 


V7 


51° 53' 12" 


59° 31' 48" 


68° 35' 



B 



11. Given a = 6, b = 8, c = 10 ; find the angles. 

12. Given a = 6, 5 = 6, c = 10 ; find the angles. 

13. Given a = 6, b = 6, c = 6 ; find the angles. 

14. Given a — 6, 6 = 5, <? = 12 ; find the angles. 

15. Given a = 2, b = V6, c = V3 — 1 ; find the angles. 

16. Given a = 2, b = V6, c = V3 + 1 ; find the angles. 

17. The distances between three cities A, B } and Care as 
follows : AB = 165 miles, A C= 72 miles, and BC= 185 miles. 
B is due east from A. In what direction is (7 from A ? What 
two answers are admissible ? 

18. Under what visual angle is an object 7 feet long seen 
by an observer whose eye is 5 feet from one end of the object 
and 8 feet from the other end ? 

19. When Formula [28] is used for finding the value of an 
angle, why does the ambiguity that occurs in Case II. not 
exist ? 

20. If the sides of a triangle are 3, 4, and 6, find the sine 
of the largest angle. 

21. Of three towns A, B, and C, A is 200 miles from B 
and 184 miles from O, B is 150 miles due north from O; how 
far is A north of C ? 



68 TRIGONOMETRY. 



~ 



§44. Area of a Triangle. 

If F denote the area of the triangle ABC (Fig. 30 or 31, 
page 50), then, by Geometry, 

F=icX CD. 

By § 10, CD = a sin B. 

Therefore, P = £ ac sin B. [33] 

And, in like manner, 

F— \ah sin C and F=% he sin A. 

That is : The area of a triangle is equal to half the product 
of two sides and the sine of the included angle. 

By Formula [33] the area of a triangle may be found directly 
when two sides and the included angle are given ; in the other 
cases the formula may be used when these parts have been 
computed. 

When the three sides of a triangle are given, as in Case IV., 
a formula for its area may be found as follows : 

By §33, sin^ = 2sinJ^Xcosi.5. 

By substituting for sin J B and cos ? B their values in terms 
of the sides given in § 43, 
2 



sin 



B = -V»(«-o)(«-4)(«-e). 



ac 



By substituting this value of sin B in [33], 

P = Vs(s-a)(s-b)(s-c). [34] 

If R denote (as in § 36) the radius of the circumscribed 
circle, we have, from § 36, 

sin B = — — • 
By substituting this value of sin B in [33], 

F = ^ c - [35] 

4R L J 



THE OBLIQUE TRIANGLE. 69 

If r denote the radius of the inscribed circle, and we divide 
the triangle into three triangles by lines from the centre of 
this circle to the vertices, the altitude of each of the three tri- 
angles is equal to r. Therefore, 

P = £r(a + b + c) = rs. [36] 

By substituting in this formula the value of F given in [34], 



r = \ 



(s-a)(s-b)(s-c) 



whence r, in [31] § 43, is equal to the radius of the inscribed 
circle. 

Exercise XVII. 



Find the area : 


1. 


Given a = 4474.5, b = 2164.5, C= 116° 30' 20". 


2. 


Given & = 21.66,' c = 36.94, A = 66° 4' 19". 


3. 


Given a = 510, = 173, B = 162° 30' 28". 


4. 


Given a = 408, 5 = 41, c = 401. 


5. 


Given a = 40, b = 13, c = 37. 


6. 


Given a = 624, b = 205, c = 445. 


7. 


Given b = 149, A = 70° 42' 30", B = 39° 18' 28". 


8. 


Given a = 215.9, c = 307.7, A = 25° 9' 31". 


9. 


Given 6 = 8, c = 5, ^4 = 60°. 


10. 


Given a =7, c = 3, A = 60°. 


11. 


Given a = 60, ^ = 40° 35' 12", area = 12; find the 


radius 


of the inscribed circle. 


12. 


Obtain a formula for the area of a parallelogram in 



terms of two adjacent sides and the included angle. 

13. Obtain a formula for the area of an isosceles trapezoid 
in terms of the two parallel sides and an acute angle. 

14. Two sides and included angle of a triangle are 2416, 
1712, and 30° ; and two sides and included angle of another 
triangle are 1948, 2848, and 150° ; find the sum of their areas. 

15. The base of an isosceles triangle is 20, and its area is 
100 -*-V3 ; find its angles. 



70 TRIGONOMETRY. 



Exercise XVIII. 

1. From a ship sailing down the English Channel the Eddy- 
stone was observed to bear N. 33° 45' W. ; and after the ship 
had sailed 18 miles S. 67°30'W. it bore N. 11° 15' E. Find 
its distance from each position of the ship. 

2. Two objects, A and B, were observed from a ship to 
be at the same instant in a line bearing N. 15° E. The ship 
then sailed north-west 5 miles, when it was found that A bore 
due east and B bore north-east. Find the distance from A 
to JB. 

3. A castle and a monument stand on the same horizontal 
plane. The angles of depression of the top and the bottom of 
the monument viewed from the top of the castle are 40° and 
80° ; the height of the castle is 140 feet. Find the height of 
the monument. 

4. If the sun's altitude is 60°, what angle must a stick make 
with the horizon in order that its shadow in a horizontal 
plane may be the longest possible ? 

5. If the sun's altitude is 30°, find the length of the longest 
shadow cast on a horizontal plane by a stick 10 feet in length. 

6. In a circle with the radius 3 find the area of the part 
comprised between parallel chords whose lengths are 4 and 5. 
(Two solutions.) 

7. A and JB, two inaccessible objects in the same horizontal 
plane, are observed from a balloon at C and from a point JD 
directly under the balloon, and in the same horizontal plane 
with A and B. If CD = 2000 yards, Z. ACD = 10° 15' 10", 
Z BCD = 6° 7' 20", A ADB = 49° 34' 50", find AB. 

8. A and B are two objects whose distance, on account of 
intervening obstacles, cannot be directly measured. At the 
summit C of a hill, whose height above the common horizontal 
plane of the objects is known to be 517.3 yards, Z ACB is 
found to be 15° 13' 15". The angles of elevation of C viewed 
from A and B are 21° 9' 18" and 23° 15' 34" respectively. 
Find the distance from A to B. 



MISCELLANEOUS PEOBLEMS. 

[Selected by permission from "Problems in Plane Trigonometry," 
prepared by Prof. C. J. White, of Harvard College, and published by 
Charles W. Sever, Cambridge.] 

1. The angular distance of any object from a horizontal 
plane, as observed at any point of that plane, is the angle 
which a line drawn from the object to the point of observa- 
tion makes with the plane. If the object observed be situated 
above the horizontal plane (that is, if it is farther from the 
earth's centre than the plane is), its angular distance from 
the plane is called its angle of elevation. If the object be 
below the plane, its angular distance from the plane is called 
its angle of depression. These angles are evidently vertical 
angles. 

If two objects are in the same horizontal plane with the 
point of observation, the angular distance of one object from 
the other is called its bearing from that object. 

If two objects are not in the same horizontal plane with 
either each other or the point of observation, we may suppose 
vertical lines to be passed through the two objects, and to 
meet the horizontal plane of the point of observation in two 
points. The angular distance of these two points is the bear- 
ing of either of the objects from the other. It may also be 
called the horizontal distance of one object from the other. 



Note. " Problems in Plane Trigonometry '' can be obtained in pam- 
phlet form of Charles W. Sever, Cambridge, Mass. 



72 



TRIGONOMETRY. 



Eight Triangles. 

2. The angle of elevation of a tower is 48° 19' 14", and the 
distance of its base from the point of observation is 95 ft. Find 
the height of the tower, and the distance of its top from the 
point of observation. 

3. From a mountain 1000 ft. high, the angle of depression 
of a ship is 77° 35' 11". Find the distance of the ship from 
the summit of the mountain. 

4. A flag-staff 90 ft. high, on a horizontal plane, casts a 
shadow of 117 ft. Find the altitude of the sun. 

5. When the moon is setting at any place, the angle at the 
moon subtended by the earth's radius passing through that 
place is 57' 3". If the earth's radius is 3956.2 miles, what is 
the moon's distance from the earth's centre ? 

6. The angle at the earth's centre subtended by the sun's 
radius is 16' 2", and the sun's distance is 92,400,000 miles. 
Find the sun's diameter in miles. 

7. The latitude of Cambridge, Mass.* is 42° 22' 49". What 
is the length of the radius of that parallel of latitude ? 

8. At what latitude is the circumference of the parallel 
half of that of the equator ? 

9. In a circle with a radius of 6.7 is inscribed a regular 
polygon of thirteen sides. Find the length of one of its sides. 

10. A regular heptagon, one side of which is 5.73, is in- 
scribed in a circle. Find the radius of the circle. 

11. A tower 93.97 ft. high is situated on the bank of a 
river. The angle of depression of an object on the opposite 
bank is 25° 12' 54". Find the breadth of the river. 



MISCELLANEOUS PKOBLEMS. 73 

12. From a tower 58 ft. high, the angles of depression of 
two objects situated in the same horizontal line with the base 
of the tower, and on the same side, are 30° 13' 18'' and 45° 
46' 14". Find the distance between these two objects. 

13. Standing directly in front of one corner of a flat-roofed 
house, which is 150 ft. in length, I observe that the horizontal 
angle which the length subtends has for its cosine VJ, and 
that the vertical angle subtended by its height has for its sine 

. What is the height of the house ? 

14. A regular pyramid, with a square base, has an edge 
150 ft. in length, and the length of a side of its base is 200 ft. 
Find the inclination of the face of the pyramid to the base. 

15. From one edge of a ditch 36 ft. wide, the angle of ele- 
vation of a wall on the opposite edge is 62° 39' 10". Find the 
length of a ladder which will reach from the point of observa- 
tion to the top of the wall. 

16. The top of a flag-staff has been broken off, and touches 
the ground at a distance of 15 ft. from the foot of the staff. 
The length of the broken part being 39 ft., find the whole 
length of the staff. 

17. From a balloon, which is directly above one town, is 
observed the angle of depression of another town, 10° 14' 9". 
The towns being 8 miles apart, find the height of the balloon. 

18. From the top of a mountain 3 miles high the angle of 
depression of the most distant object which is visible on the 
earth's surface is found to be 2° 13' 50". Find the diameter 
of the earth. 

19. A ladder 40 ft. long reaches a window 33 ft. high, on 
one side of a street. Being turned over upon its foot, it 
reaches another window 21 ft. high, on the opposite side of 
the street. Find the width of the street. 



74 TRIGONOMETRY. 



20. The height of a house subtends a right angle at a win- 
dow on the other side of the street ; and the elevation of the 
top of the house, from the same point, is 60°. The street is 
30 ft. wide. How high is the house ? 

21. A lighthouse 54 ft. high is situated on a rock. The 
elevation of the top of the lighthouse, as observed from a ship, 
is 4° 52', and the elevation of the top of the rock is 4° 2'. 
Find the height of the rock, and its distance from the ship. 

22. A man in a balloon observes the angle of depression of 
an object on the ground, bearing south, to be 35° 30'; the 
balloon drifts 2 J miles east at the same height, when the angle 
of depression of the same object is 23° 14'. Find the height 
of the balloon. 

23. A man standing south of a tower, on the same horizon- 
tal plane, observes its elevation to be 54° 16' : he goes east 
100 yds., and then finds its elevation is 50° 8'. Find the 
height of the tower. 

24. The elevation of a tower at a place A south of it is 30° ; 
and at a place B, west of A, and at a distance of a from it, the 
the elevation is 18°. Show that the height of the tower is 

V5-1 



the tangent of 18° being 



V(2+2V5) V(l() + 2V5) 

25. A pole is fixed on the top of a mound, and the angles 
of elevation of the top and the bottom of the pole are 60° and 
30°. Prove that the length of the pole is twice the height of 
the mound. 

26. At a distance (a) from the foot of a tower, the angle 
of elevation (^4) of the top of the tower is the complement of- 
the angle of elevation of a flag-staff on top of it. Show that 
the length of the staff is 2 a cot 2 A. 

27. A line of true level is a line every point of which is 
equally distant from the centre of the earth. A line drawn 



MISCELLANEOUS PROBLEMS. 75 

tangent to a line of true level at any point is a line of appar- 
ent level. If at any point both these lines be drawn, and 
extended one mile, find the distance they are then apart. 

28. In Problem 2, determine the effect upon the computed 
height of the tower, of an error in either the angle of eleva- 
tion or the measured distance. 

Oblique Triangles. 

29. To determine the height of an inaccessible object situ- 
ated on a horizontal plane, by observing its angles of elevation 
at two points in the same line with its base, and measuring 
the distance of these two points. 

30. The angle of elevation of an inaccessible tower, situated 
on a horizontal plane, is 63° 26' ; at a point 500 ft. farther 
from the base of the tower the elevation of its top is 32° 14'. 
Find the height of the tower. 

31. A tower is situated on the bank of a river. From the 
opposite bank the angle of elevation of the tower is 60° 13', 
and from a point 40 ft. more distant the elevation is 50° 19'. 
Find the breadth of the river. 

32. A ship sailing north sees two lighthouses 8 miles apart, 
in a line due west ; after an hour's sailing, one lighthouse bears 
S.W., and the other S.S.W. Find the ship's rate. 

33. To determine the height of an accessible object situated 
on an inclined plane. 

34. At a distance of 40 ft. from the foot of a tower on an 
inclined plane, the tower subtends an angle of 41° 19' ; at a 
point 60 ft. farther away, the angle subtended by the tower is 
23° 45'. Find the height of the tower. 

35. A tower makes an angle of 113° 12' with the inclined 
plane on which it stands ; and at a distance of 89 ft. from its 
base, measured down the plane, the angle subtended by the 
tower is 23° 27'. Find the height of the tower. 



76 TRIGONOMETRY. 



36. From the top of a house 42 ft. high, the angle of eleva- 
tion of the top of a pole is 14° 13' ; at the bottom of the house 
it is 23° 19'. Find the height of the pole. 

37. The sides of a triangle are 17, 21, 28 ; prove that the 
length of a line bisecting the greatest side and drawn from 
the opposite angle is 13. 

38. A privateer, 10 miles S.W. of a harbor, sees a ship sail 
from it in a direction S. 80° E., at a rate of 9 miles an hour. 
In what direction, and at what rate, must the privateer sail 
in order to come up with the. ship in 1-J- hours ? 

39. A person goes 70 yds. up a slope of 1 in 3-J- from the 
edge of a river, and observes the angle of depression of an ob- 
ject on the opposite shore to be 2J°. Find the breadth of the 
river. 

40. The length of a lake subtends, at a certain point, an 
angle of 46° 24', and the distances from this point to the two 
extremities of the lake are 346 and 290 ft. Find the length 
of the lake. 

41. Two ships are a mile apart. The angular distance of 
the first ship from a fort on shore, as observed from the second 
ship, is 35° 14' 10" ; the angular distance of the second ship 
from the fort, observed from the first ship, is 42° 11' 53". Find 
the distance in feet from each ship to the fort. 

42. Along the bank of a river is drawn a base line of 500 
feet. The angular distance of one end of this line from an 
object on the opposite side of the river, as observed from the 
other end of the line, is 53° ; that of the second extremity 
from the same object, observed at the first, is 79° 12'. Find 
the perpendicular breadth of the river. 

43. A vertical tower stands on a declivity inclined 15° to 
the horizon. A man ascends the declivity 80 ft. from the base 
of the tower, and finds the angle then subtended by the tower 
to be 30°. Find the height of the tower, 



MISCELLANEOUS PROBLEMS. 77 

44. The angle subtended by a tower on an inclined plane 
is, at a certain point, 42° 17' ; 325 ft. farther down, it is 21° 
47'. The inclination of the plane is 8° 53'. Find the height 
of the tower. 

45. A cape bears north by east, as seen from a ship. The 
ship sails northwest 30 miles, and then the cape bears east. 
How far is it from the second point of observation ? 

46. Two observers, stationed on opposite sides of a cloud, 
observe its angles of elevation to be 44° 56' and 36° 4'. Their 
distance from each other is 700 ft. What is the linear height 
of the cloud ? 

47. From a point B at the foot of a mountain, the eleva- 
tion of the top A is 60°. After ascending the mountain one 
mile, at an inclination of 30° to the horizon, and reaching a 
point C, the angle ACB is found to be 135°. Find the height 
of the mountain in feet. 

48. -From a ship two rocks are seen in the same right line 
with the ship, bearing N. 15° E. After the ship has sailed 
northwest 5 miles, the first rock bears east, and the second 
northeast. Find the distance between the rocks. 

49. From a window on a level with the bottom of a steeple 
the elevation of the steeple is 40°, and from a second window 
18 ft. higher the elevation is 37° 30'. Find the height of the 
steeple. 

50. To determine the distance between two inaccessible 
objects by observing angles at the extremities of a line of 
known length. 

51. Wishing to determine the distance between a church A 
and a tower B, on the opposite side of a river, I measure a 
line CD along the river {C being nearly opposite A), and ob- 
serve the angles ACB, 58° 20' ; A CD, 95° 20' ; ADB, 53° SO'; 
BDC } 98° 45'. CD is 600 ft. What is the distance required ? 



78 TRIGONOMETRY. 

52. Wishing to find the height of a summit A, I measure 
a horizontal base line CD, 440 yds. At C, the elevation of A 
is 37° 18', and the horizontal angle between D and the sum- 
mit is 76° 18' ; at D, the horizontal angle between and the 
summit is 67° 14'. Find the height. 

53. A balloon is observed from two stations 3000 ft. apart. 
At the first station the horizontal angle of the balloon and the 
other station is 75° 25', and the elevation of the balloon is 18°. 
The horizontal angle of the first station and the balloon, meas- 
ured at the second station, is 64° 30'. Find the height of the 
balloon. 

54. Two forces, one of 410 pounds, and the other of 320 
pounds, make an angle of 51° 37'. Find the intensity and the 
direction of their resultant. 

55. An unknown force, combined with one of 128 pounds, 
produces a resultant of 200 pounds, and this resultant makes 
an angle of 18° 24' with the known force. Find the intensity 
and direction of the unknown force. 

56. At two stations, the height of a kite subtends the same 
angle (A). The angle which the line joining one station and 
the kite subtends at the other station is B ; and the distance 
between the two stations is a. Show that the height of the 
kite is \ a sin A sec B. 

57. Two towers on a horizontal plane are 120 ft. apart. A 
person standing successively at their bases observes that the 
angular elevation of one is double that of the other ; but, when 
he is half-way between them, the elevations are complementary. 
Prove that the heights of the towers are 90 and 40 ft. 

58. To find the distance of an inaccessible point C from 
either of two points A and B, having no instruments to meas- 
ure angles. Prolong CA to a, and CB to b, and join AB, Ab, 
and Ba. Measure AB, 500; a A, 100; aB, 560; bB, 100; 
and Ab, 550. 



MISCELLANEOUS PEOBLEMS. 79 

59. Two inaccessible points A and B, are visible from D, 
but no other point can be found whence both are visible. 
Take some point C, whence A and D can be seen, and meas- 
ure CD 200 ft. ; ADC, 89° ; ACD, 50° 30'. Then take some 
point E, whence D and B are visible, and measure DE, 200 ; 
BDE, 54° 30' ; BED, 88° 30'. At D measure ADD, 72° 30'. 
Compute the distance AB. 

60. To compute the horizontal distance between two inac- 
cessible points A and B, when no point can be found whence 
both can be seen. Take two points C and D, distant 200 yds. 
so that A can be seen from C, and B from D. From C meas- 
ure CF, 200 yds. to E, whence A can be seen ; and from D 
measure DE, 200 yds. to E, whence B can be seen. Measure 
AFC, 83°; ACD, 53° 30' ; ACE, 54° 31' ; BDE, 54° 30'; 
BDC, 156° 25' ; DEB, 88° 30'. 

61. A column in the north temperate zone is east-southeast 
of an observer, and at noon the extremity of its shadow is 
northeast of him. The shadow is 80 ft. in length, and the 
elevation of the column, at the observer's station, is 45°. Find 
the height of the column. 

62. From the top of a hill the angles of depression of two 
objects situated in the horizontal plane of the base of the hill 
are 45° and 30° ; and the horizontal angle between the two 
objects is 30°. Show that the height of the hill equals the 
distance between the objects. 

63. Wishing to know the breadth of a river from A to B, 
I take AC, 100 yds. in the prolongation of BA, and then take 
CD, 200 yds. at right angles to AC. The angle BDA is 37° 
18' 30". Find AB. 

64. The sum of the sides of a triangle is 100. The angle 
at A is double that of B, and the angle at B is double that 
at C Determine the sides. 



80 



TRIGONOMETRY. 



65. If sin 2 ^. + 5 cos 2 A = 3, find A. 

66. If sin 2 A = mcosA — n, find cos A. 

67. Given sin A = m sin i?, and tan A = n tan i?, find sin A 
and cos J3. 

68. If tan 2 J. + 4 sin 2 ^ = 6, find A. 

69. If sin A = sin 2^4, find A. 

70. If tan 2 ^4 = 3 tan A, find X 

71. Prove that tan 50° -f cot 50° = 2 sec 10°. 

72. Given a regular polygon of n sides, and calling one of 
them a, find expressions for the radii of the inscribed and the 
circumscribed circle in terms of n and a. 

If P, H, D be the sides of a regular inscribed pentagon, 
hexagon, and decagon, prove P 2 = H 2 -\- D 2 . 

Areas. 

73. Obtain the formula for the area of a triangle, given two 
sides b, c, and the included angle A. 

74. Obtain the formula for the area of a triangle, given two 
angles A and B, and included side c. 

75. Obtain the formula for the area of a triangle, given the 
three sides. 

76. If a is the side of an equilateral triangle, its area is: 
a 2 V3 



77. Two consecutive sides of a rectangle are 52.25 ch. and 
38.24 ch. Find its area. 

78. Two sides of-a parallelogram are 59.8 ch. and 37.05 ch., 
and the included angle is 72° 10'. Find the area. 

79. Two sides of a parallelogram are 15.36 ch. and 11.46 
ch., and the included angle is 47° 30'. Find its area. 



MISCELLANEOUS PKOBLEMS. 81 

80. Two sides of a triangle are 12.38 ch. and 6.78 ch., and 
the inclined angle is 46° 24'. Find the area. 

81. Two sides of a triangle are 18.37 ch. and 13.44 ch., and 
they form a right angle. Find the area. 

82. Two angles of a triangle are 76° 54' and 57° 33' 12", 
and the included side is 9 ch. Find the area. 

83. Two sides of a triangle are 19.74 ch. and 17.34 ch. 
The first bears N. 82° 30' W. ; the second, S. 24° 15' E. Find 
the area. 

84. The three sides of a triangle are 49 ch., 50.25 ch., and 
25.69 ch. Find the area. 

85. The three sides of a triangle are 10.64 ch., 12.28 ch., 
and 9 ch. Find the area. 

86. The sides of a triangular field, of which the area is 14 
acres, are in the ratio of 3, 5, 7. Find the sides. 

87. In the quadrilateral ADCD'we have AD, 17.22 ch. ; 
AD, 7.45 ch. ; CD, 14.10 ch. ; DC, 5.25 ch. ; and the diago- 
nal AC, 15.04 ch. Required the area. 

88. The diagonals of a quadrilateral are a and b, and they 
intersect at an angle D. The area of the quadrilateral is 
\ ab sin D. 

89. The diagonals of a quadrilateral are 34 and 56, inter- 
secting at an angle of 67°. Find the area. 

90. The diagonals of a quadrilateral are 75 and 49, inter- 
secting at an angle of 42°. Find the area. 

91. The area of a regular polygon of n sides, of which one 

• na 2 ,180° 

is a, is — cot- 

4 n 

92. One side of a regular pentagon is 25. Find the area. 

93. One side of a regular hexagon is 32. Find the area. 



82 TRIGONOMETRY. 



94. One side of a regular decagon is 46. Find the area. 

95. Find the area of a circle whose circumference is 74 ft. 

96. Find the area of a circle whose radius is 125 ft. 

97. In a circle with a diameter of 125 ft. find the area of a 
sector with an arc of 22°. 

98. In a circle with a radius of 44 ft. find the area of a 
sector with an arc of 25°. 

99. In a circle with a diameter of 50 ft. find the area of a 
segment with an arc of 280°. 

100. Find the area of a segment (less than a semicircle), of 
which the chord is 20, and the distance of the chord from the 
middle point of the smaller arc is 2. 

101. If r is the radius of a circle, the area of a regular cir- 

180° 

cumscribed polygon of n sides is nr 2, tan 

n 

Q£>f\0 

The area of a regular inscribed polygon is - r 2 sin 

102. If a is a side of a regular polygon of n sides, the area 

of the inscribed circle is — cot 2 

4 n 

The area of the circumscribed circle is — esc 2 

4 n 

103. The area of a regular polygon inscribed in a circle is 
to that of the circumscribed polygon of the same number of 
sides as 3 to 4. Find the number of sides. 



104. The area of a regular polygon inscribed in a circle is 
a geometric mean between the areas of an inscribed and a cir- 
cumscribed regular polygon of half the number of sides. 

105. The area of a circumscribed regular polygon is an har- 
monic mean between the areas of an inscribed regular polygon 
of the same number of sides, and of a circumscribed regular 
polygon of half that number. 



MISCELLANEOUS PROBLEMS. 83 

106. The perimeter of a circumscribed regular triangle is 
double that of the inscribed regular triangle. 

107. The square described about a circle is four-thirds the 
inscribed dodecagon. 

108. Two sides of a triangle are 3 and 12, and the included 
angle is 30°. Find the hypotenuse of an isosceles right tri- 
angle of equal area. 

Plane Sailing. 

109. Plane Sailing is that branch of Navigation in which 
the surface of the earth is considered a plane. The problems 
which arise are therefore solved by the methods of Plane 
Trigonometry. 

The following definitions will explain the technical terms 
which are employed : 

The difference of latitude of two places is the arc of a merid- 
ian comprehended between the parallels of latitude passing 
through those places. 

The departure between two meridians is the arc of a par- 
allel of latitude comprehended between those meridians. It 
evidently diminishes as the distance from the equator at which 
it is measured increases. 

When a ship sails in such a manner as to cross successive 
meridians at the same angle, it is said to sail on a rhumb-line. 
The constant angle which this line makes with the meridians 
is called the course, and the distance between two places is 
measured on a rhumb-line. 

If we neglect tne curvature of the earth, and consider the 
distance, departure, and difference of latitude of two places to 
be straight lines, lying in one plane, they will form a right 
triangle, called the triangle of plane sailing. If ABB be a 
plane triangle, right-angled at B, and AB represent the dif- 
ference of latitude of A and B, BAB will be the course from 



84 TRIGONOMETRY. 



A to B, AB the distance, and DB the departure, measured 
from B, between the meridian of A and that of B. 

110. Taking the earth's equatorial diameter to be 7925.6 
miles, find the length in feet of the arc of one minute of a 
great circle.* 

111. A ship sails from latitude 43° 45' S., on a course N. 
by E., 2345 miles. Find the latitude reached, and the 
departure made. 

112. A ship sails from latitude 1° 45' N., on a course S.E. 
by E., and reaches latitude 2° 31' S. Find the distance, and 
the departure. 

113. A ship sails from altitude 13° 17' S., on a course N.E. 
by E. } E., until the departure is 207 miles. Find the dis- 
tance, and the latitude reached. 

114. A ship sails on a course between S. and E., 244 
miles, leaving latitude 2° 52' S., and reaching latitude 5° 8' 
S. Find the course, and the departure. 

115. A ship sails from latitude 32° 18' N., on a course be- 
tween N. and W., making a distance of 344 miles, and a 
departure of 103 miles. Find the course, and the latitude 
reached. 

116. A ship sails on a course between S. and E., making 
a difference of latitude 136 miles, and a departure 203 miles. 
Find the distance, and the course. 

117. A ship sails due north 15 statute miles an hour, for 
one day. What is the distance, in a straight line, from the 
point left to the point reached ? (Take earth's radius, 3962.8 
statute miles.) 

* The length of the arc of one minute of a great circle of the earth 
is called a geographical mile, or a knot. In the following problems, this 
is the distance meant by the term "mile," unless otherwise stated. 



miscellaneous problems. 85 

Parallel and Middle Latitude Sailing. 

118. The difference of longitude of two places is the angle 
at the pole made by the meridians of these two places ; or, it 
is the arc of the equator comprehended between these two 
meridians. 

119. In Parallel Sailing, a vessel is supposed to sail on a 
parallel of latitude ; that is, either due east or due west. 
The distance sailed is, in this case, evidently the departure 
made ; and the difference of longitude made depends on the 
solution of the following problem : 

120. Given the departure between any two meridians at 
any latitude, find the angle which those meridians make, 
or the difference of longitude of any point on one meridian 
from any point on the other. (The earth is considered to 
be a perfect sphere, and the solution depends on simple 
geometric and trigonometric principles. Of. Problem 7.) 
The solution gives the following formula : 

Diff. long. = depart. X sec. lat. 

121. A ship in latitude 42° 16' N., longitude 72° 16' W., 
sails due east a distance of 149 miles. What is the position 
of the point reached ? 

122. A ship in latitude 44° 49' S., longitude 119° 42' E., sails 
due west until it reaches longitude 117° 16' E. Find the dis- 
tance made. 

123. In Middle Latitude Sailing, the departure between two 
places, not on the same parallel of latitude, is considered to 
be, approximately, the departure between the meridians of 
those places, measured on that parallel of latitude which lies 
midway between the parallels of the two places. Except 
in very high latitudes or excessive runs, such an assumption 
produces no great error. By the formula of Art. 120, then, 
we shall have — 

Diff. long. = depart. X sec. mid. lat. 



86 TRIGONOMETRY. 



124. A ship leaves latitude 31° 14' N., longitude 42° 19' 
W., and sails E.N.E. 325 miles. Find the position reached. 

125. Find the bearing and distance of Cape Cod from 
Havana. (Cape Cod, 42° 2' N., 70° 3' W. ; Havana, 23° 9' N., 
82° 22' W.) 

126. Leaving latitude 49° 57' N., longitude 15° 16' W., a 
ship sails between S. and W. till the departure is 194 miles, 
and the latitude is 47° 18' N. Find the course, distance, 
and longitude reached. 

127. Leaving latitude 42° 30' N., longitude 58° 51' W., a 
ship sails S.E. by S. 300 miles. Find the position reached. 

128. Leaving latitude 49° 57' N., longitude 30° W., a ship 
sails S. 39° W., and reaches latitude 47° 44' N. Find the 
distance, and longitude reached. 

129. Leaving latitude 37° N., longitude 32° 16' W., a ship 
sails between N. and W. 300 miles, and reaches latitude 41° 
N. Find the course, and longitude reached. 

130. Leaving latitude 50° 10' S., longitude 30° E., a ship 
sails E.S.E., making 160 miles' departure. Find the distance, 
and position reached. 

131. Leaving latitude 49° 30' N., longitude 25° W., a ship 
sails between S. and E. 215 miles, making a departure of 167 
miles. Find the course, and position reached. 

132. Leaving latitude 43° S., longitude 21° W., a ship sails 
273 miles, and reaches latitude 40° 17' S. What are the two 
courses and longitudes, either one of w T hich will satisfy the 
data? 

133. Leaving latitude 17° N., longitude 119° E., a ship sails 
219 miles, making a departure of 162 miles. What four sets 
of answers do we get ? 



MISCELLANEOUS PROBLEMS. 87 

134. A ship in latitude 30° sails due east 360 statute miles. 
What is the shortest distance from the point left to the point 
reached ? 

Solve the same problem for latitude 45°, 60°, etc. 



Traverse Sailing. 

135. Traverse Sailing is the application of the principles of 
Plane and Middle Latitude Sailing to cases when the ship 
sails from one point to another on two or more different 
courses. Each course is worked up by itself, and these 
independent results are combined, as may be seen in the 
solution of the following example : 

136. Leaving latitude 37° 16' S., longitude 18° 42' W., a 
ship sails N.E. 104 miles, then N.N.W. 60 miles, then W. by 
S. 216 miles. Find the position reached, and its bearing and 
distance from the point left. 

We have, for the first course, difference of latitude 73.5 N., 
departure 73.5 E. 

We have, for the second course, difference of latitude, 55.4 
N., departure 23 W. 

We have, for the third course, difference of latitude 42.1 S., 
departure 211.8 W. 

On the whole, then, the ship has made 128.9 miles of 
north latitude, and 42.1 miles of south latitude. The place 
reached is therefore on a parallel of latitude 86.8 miles to 
the north of the parallel left ; that is, in latitude 35° 49'. 2 S. 

The departure is, in the same way, found to be 161.3 miles 
W. ; and the middle latitude is 36° 32'.6. With these data, 
and the formula of Art. 126, we find the difference of longi- 
tude to be 201 miles, or 3° 21' W. Hence the longitude 
reached is 22° 3' W. 

With the difference of latitude 86.8 miles, and the depart- 
ure 161.3 miles, we find the course to be N. 61° 43' W., and 



TRIGONOMETRY. 



the distance 183.2 miles. The ship has reached the same 
point that it would have reached, if it had sailed directly on 
a course N. 61° 43' W., for a distance of 183.2 miles. 

137. A ship leaves Cape Cod (Ex. 125), and sails S.E. by 
S. 114 miles, N. by E. 94 miles, W.N.W. 42 miles. Solve as 
in Art. 139. 

138. A ship leaves Cape of Good Hope (latitude 34° 22' S., 
longitude 18° 30' E.), and sails N.W. 126 miles, N. by E. 84 
miles, W.S.W. 217 miles. Solve as in Ex. 136. 



EXAMINATION PAPERS.* 



PLANE TRIGONOMETRY. 

I. 

{Harvard College, Admission. June, 1881. Time, 1J hours.) 

1. Define a logarithm. "What is the logarithm of i in the 
system of which 16 is the base ? Find the logarithm of 25 in 
the same system. 

2. Compute the value of \ i 4 ^ x q Q0651 h 7 logarithms. 

3. Find the functions of 127° 10' from your trigonometric 
tables. 

4. Prove the formula 

(cos A - cos Bf + (sin A - sin Bf = 4 sin* A 



2 

5. Two sides of a triangle are 243 feet and 188 feet, and 
the angle opposite the second side is 42° 20'. Solve the tri- 
angle completely. 

6. A pine tree growing on the side of a mountain, which is 
inclined to the horizontal at an angle of 20°, is broken by the 
wind but not severed at a distance of 40 feet from the ground. 
The top falls toward the foot of the mountain, and strikes the 
ground 50 feet from the base of the tree ; find the height of 
the tree. 

* Note. In these papers, as in many text-books, the Greek letters a 
(alpha), (3 (bayta), y (gamma), 6 (delta), 6 (ihayta), <j> (phee), are occasionally 
used to denote angles. 



90 



TRIGONOMETRY. 



II. 
{Harvard College, Admission. June, 1882. Time, 1J hours.) 

1. Explain the reason of the rule for finding the character- 
istic (or integral part) of the logarithm of a number. 

Show that (according to this rule) the mantissa (or frac- 
tional part) is always positive. 

In what cases is the logarithm, as a whole, positive, and in 
what cases negative ? 

Thus, state clearly the value of the logarithm of 36,270 ; of 
0.003627. What decimal must be added to the latter loga- 
rithm to produce the logarithm of 0.01 ? 

2. Find the time required to increase a sum of money a 
hundred fold, at ten per cent per annum, compound interest, 
payable yearly. 

3. Find the formulas for the trigonometric functions of 

90°+ a. 

4. Find by the tables the logarithms of the trigonometric 
functions of 290° 38' (marking the signs). 

5. An observer from a ship saw two headlands. The first 
bore E.N.E. (i.e. 67° 30' from N. towards E.), and the second 
N.W. by N. (i.e. 33° 45' from N. towards W.). After he had 
sailed 16.25 miles N. by W. (i.e. 11° 15' from N. towards W.J, 
the first headland bore due E., and the second N.W. by W. 
(i.e. 56° 15' from N. towards W.). Find the direction and dis- 
tance of the second headland from the first. 



6. Prove the formulas 



eosa-f-cos/J 
sin# 



^ = -tanKa + £)tanHa-j8), 



2tan£0 



l + tan 2 -i-0 



EXAMINATION PAPERS. 91 

III. 
(Harvard College, Freshman Examination. April, 1879. Time, 3 hours.) 

1. Prove the relations between the sine, cosine, and tan- 
gent of 90° + and the functions of <f>. Draw a figure for the 
case where <£ is obtuse, and show that the proof still holds 
good. 

Confirm your results by means of the formulas for the sine 
and cosine of the sum of the two angles. 

2. Deduce formulas for the sine, cosine, and tangent of 2a 
and -J a, in terms of functions of a. 

3. Prove the formula : 

cos (a -f- /?) sin /? — cos (a -f- y) sin y = 
sin (a + /?) cos ft — sin (a -f- y) cosy. 

4. Prove that in any triangle 

a 2 = b 2 + c 2 — 2 be cos A. 

5. Deduce the formulas for the tangents of the half angles 
of a triangle, in terms of the sides. 

6. Solve the triangles : 

(7=35°, a = 500, * = 250, 

B = 22° 22', a = 67.06, b = 60.03. 

7. The Delta measures 241 yds. on Cambridge St. and 115 
yds. on Quincy St., and the angle between these streets is 
88° 52'. Find the other angles of the Delta. 

8. Find the area of the Delta. 

9. A person travelling east in a railroad train observes a 
tower situated south of a station A, and on the same horizontal 
plane with the railroad. At the station B he finds that the 
distance of the tower is 2 miles ; and at C, 3 miles from B, its 
distance is 4 miles. Find the distance of the tower from A. 



92 



TRIGONOMETRY. 



IV. 

{Harvard College, Freshman Examination. April, 1880. Time, 3 hours.) 

1. Deduce the formula sin (a -f /?) = , drawing the figure 

for the case in which a is in the second quadrant and a + ft 
in the third. 

2. Deduce the formulas for cos 2 a, sin J a, and cos|-a, in 
terms of functions of a. 

3. Prove the theorem of the sines. 

4. From the fundamental formulas deduce the formula 



tzn±-(A + B)_a + h 
tan \{A-B) a-b 



5. Prove that 



\arptyj _ 2 _ t an a tan /? — tan /? tan y — tan y tan a. 
cos a cos/8 cosy 

6. In a triangle ^ = 4° 13.4' and a = 2001, give all the 
solutions in the following cases : 

(1) 5 = 150, 

(2) 5 = 200, 

(3) 5 = 2001. 

7. A, B, and Care the corners of a triangular field. A is 
40 ft. W. of B and 400 ft. S.W. of C. What is the area of 
the field ? "What is the length of the fence which encloses it ? 

8. From two corners of the Delta, A and B, lines which 
make angles of 19° 52' and 57° 32' respectively with the side 
AB meet directly under Memorial Hall tower. The length 
of AB is 345.1 ft., and the angle of elevation of the tower at 
A is 32° 26'. Find the height of the tower, and its angle of 
elevation at B. 



EXAMINATION PAPERS. 93 

V. 

{Harvard College, Freshman Examination. April, 1881. Time, 3 hours.) 

1. Deduce the formulas which connect the functions of 
(90° + <£) and<£. 

2. Prove the fundamental formula for cos (a -f- /?)• 

3. From the formula just found, obtain three values for 
cos 2 a. 

4. Find the values of sin- and cos -, in terms of cos a. 

2 2 

5. Prove the formula 

(cos A - cos Bf + (sin A - sin Bf = 4 sin 2 ^=-^- 

6. Solve the following triangles : 

b = 2434, = 1881, c = 42° 22 r . 

a = 0.00543, c = 0.07003, a = 4° 27'. 

7. The sides of a triangle are 715, 541, and 368 ; find one 
angle and the area. 

8. The height of Memorial Hall tower is 190 feet. From its 
top the angles of depression of the corners of the Delta which 
lie on Cambridge St. are 57° 44' and 16° 59', and the angle 
subtended by the line joining these corners is 99° 30'. Find 
the length of the Delta on Cambridge St. 



VI. 

(Harvard College, Freshman Examination. April, 1882. Time, 3 hours.) 

1. Obtain the formulas which connect the sine, cosine, and 
tangent of (180° -f </>) with the functions of <£. 

2. Assuming the formulas for the sine and cosine of the 
sum of two angles, prove that 

(1) tan (« + £)= 

(2) sin^a= V|(l — cos a). 



94 TRIGONOMETRY. 



3. Find all the values of x, between 0° and 360°, which will 
satisfy the equations 

(1) tan x = 2 sin 2 x, 

(2) (sin x -f- cos x) 2 == 2 sin 2x. 

4. The length of each side of a regular dodecagon is 24 
feet ; find the radius of the inscribed circle and the area of the 
polygon. 

5. In a certain triangle, a = 20, B = 3° 24', C= 85° 31'. 
Find c by aid of the table containing the values of /S and T. 

6. Prove the Theorem of Sines, and solve the triangles 

(1) 5-468, c = 327, C=34°15'; 

(2) a= 0.003641, c = 0.08091, A = 5°20 r . 

7. Given a 2 = b 2 -f- c 2 — 2 be cos A ; obtain the formula 



sin $A = Y 



(s-b)(s-c) 
be 



The sides of a triangle are a = 2408, b = 2028, c = 1884 ; 
find the angle A. 

8. Hingham is 12 miles south-east of Boston ; Quincy is 5f 
miles west of Hingham. How far is Quincy from Boston ? 

9. A person ascending Memorial Hall tower stops to rest 
at a window, and notices that the angle of elevation of the 
vane on Appleton Chapel is 3° 34'. When he reaches the top 
of the tower, 190 feet above the ground, he finds that the 
height of the Chapel subtends an angle of 13° 9'. The hori- 
zontal distance between the two towers being 492 feet, find 
the height of Appleton Chapel and the distance of the window 
above the ground. 



EXAMINATION PAPERS. 95 

VII. 

{Cambridge, Eng. 2nd Previous Exam., Dec. 7, 1876. Time, 2\ hours.) 

1.* Assuming that the angle subtended at the centre of any 
circle by an arc equal to its radius is a constant angle, show 
that any angle may be expressed by the fraction ■ — — — , the 
constant angle being taken as the unit. 

Find the length of the arc subtended by an angle of 60° in 
a circle whose radius is 3 feet. 

2. The sine of a certain angle is \ ; find the other trigono- 
metrical ratios of the angle. 

3. Trace the change in sign and magnitude in the tangent 
of an angle, as the angle increases from 0° to 360°. 

4. Find, by a geometrical construction, the cosine of 60° 
and of 45°, and deduce the value of cos 3360° and cos 2565°. 

5. Prove the formulas : 

(1) sin {A — B) = sin A cos B — cos A sin B, 

/o\ o a 1 — tan 2 J. 

(2) cos2^L = - — > 

v J 1 + tan 2 ^ 

(3) sin2 ^ + sin4 ^ = tan3A 

v J cos 2 ,4 -f cos 4 J. 

6. Express the cosine of half an angle in terms of the sine 
of the angle, and explain the double sign. 

Employ the formula to find the value of cos 75°, having 
given sin 150° = \. 

7. If A, B, C be the angles of a triangle, and a, b, c the 
sides respectively opposite to them, show that 

where s = one-half the sum of the sides. 

8. Find the greatest angle in a triangle whose sides are 7 
feet, 8 feet, and 9 feet. 

*For aid in solving this and similar questions, see Wentworth & Hill's 
Tables, pages xi and xii. 



COS 



96 TRIGONOMETRY. 



VIII. 

{Cambridge, Eng. 2nd Previous Exam., Dec. 7, 1877. Time, 2£ hours.) 

1. Define the cosine, cotangent, and cosecant of an angle, 
and prove that these ratios remain unchanged so long as the 
angle is the same. 

Find the value of these three ratios for an angle of 45°. 

2. Prove the formulas : 

(1) sin A = VI - cos 2 A, 

(2) cos^L- j 



Vl-f-tan 2 ^ 
If sec A = V2, find tan A. Ans. 1. 

3. Prove that 
sin (90° + A) = cos .4, and cos (90° + A) = — sin A. 

Hence show that cos (180° + A) = — cos A. 

4. Show that cos 2 A tan 2 ^4 -f- sin 2 A cot 2 A = 1. 

5. Prove that cos (^4 -f- B) = cos A cos B — sin A sin B. 
Hence show that 

cos(J. + ^+^) ^ cot ^ cot ^ cot ^_ cot ^_ cot ^_ cota 
sin A sin B sin G 

6. Given that sin^A = f, find the value of tan A. 

Ans. -3j£. 

7. Prove that the sides of any plane triangle are propor- 
tional to the sines of the angles opposite to these sides. 

If 2s = the sum of the three sides (a, b, c) of a triangle, and 
if A be the angle opposite to the side a, prove that 

. A 2 , 

sin A = — Vs(s — a)(s — b) (s — c). 

8. Prove that in any plane triangle 



tan J(^L - B) = — | cot \Q. 



EXAMINATION PAPERS. 97 

9. If the side a and the angles A and B of a triangle be 

known, prove that the side b may be found by means of the 

formula, 

log b = log a + log sin B — log sin A. 

Find b, having given that a = 1000 yards, J. = 50°, .£=64°. 

Ans. 1173.29 yards. 

10. The minute-hand of a clock is 3 feet 6 inches in length ; 
find how far its point will move in a quarter of an hour, it 
being assumed that n = - 2 ^. Ans. 5 feet 6 inches. 



IX. 

(Cambridge, Eng. 2nd Previous Exam., Dec. 10, 1878. Time, 2\ hours.) 

1. Define sine, cotangent; and prove that sin 2 ^4-f cos 2 ^4=1. 
Express the other trigonometrical ratios in terms of the cosine. 

2. What is meant by the circular measure of an angle? 
How is the number of degrees in an angle found from its cir- 
cular measure ? How many degrees are in the unit of circular 
measure ? 

3. Prove that 

(1) sin (180°+^) = -sin A, 

(2) tan (90°+^) = - cot A 
What is the use of these equations ? 

4. Find the general form of all the angles whose sine is the 
same as sin 0. 

Write down the sines of all the angles which are multiples 
of 30° and less than 360°. 



TRIGONOMETRY. 



5. Prove the following relations : 

(1) cos ( A — B) = cos A cos B -f sin A sin B, 

(2) sin^ + sin^ = 2sin|(^-[-^) cos±(A-B), 

/Q n 2 i 1 — cos 2 ^4. 

(3) tan 2 A = — -■ 

v J 1-f cos2^4 

6. Find cos 30°, tan 45°, sin 15°. 

7. If tan A -f- sec ^. = 2, prove that sin ^4 = -| when ^ is less 
than 90°. 

If sin A = %, prove that tan A -J- sec J. = 3 when ^4 is less 
than 90°. 

8. Prove that cos 3 A = 4 cos 3 A — 3 cos A, and find tan 3 A 
in terms of tan A. 

9. In a triangle A,B } C, whose sides are respectively a, b, c, 
prove that 

(i) ■ + A=^fe=a£E3, 

•p\ sin ^4 sini? sin (7 

a & c 

10. Solve a triangle, having given two sides and the angle 
opposite one of them. 

Find A, B, b, having given a = 25, c = 24, C— 65° 59'. 



X. 

(Cambridge, Eng. 2nd Previous Exam., Bee. 10, 1879. Time, 2\ hours.) 

1. Define 1°. Assuming that ^ is the circular measure of 
two right angles, express the angle A° in circular measure. 

2. Define the sine, secant, and cotangent of an angle, and 
express any two of these ratios in terms of the third. 

Find the trigonometrical ratios of the angle whose cosine 
is 4. 



EXAMINATION PAPERS. 99 

3. Prove that 

(1) cos (180° + A) = cos (180° -A), 

(2) tan (90° + ^) = cot(180°-^). 

4. Express the cosine of the difference of two angles in 
terms of the sines and cosines of these angles. 

Prove that 

*tan -1 # -f- tan -1 y = tan -1 ^ • 

l-xy 

5. Prove the formulas : 

0C I ' 1/ oc — XI 

(1) cos x-\- cosy = 2 cos — i^ cos — — £, 



(2) sin \x -f cos-Jo; = ± Vl + sin ic, 

(3) sin:r(2 cosa; — 1) = 2 sin -J- a: cosfa?. 

6. Trace the changes in sign and magnitude of 

2sin(9-sin2(9 
2sin0-fsin20' 

as changes from to 2 ir. 

7. Express the cosine of any angle of a triangle in terms of 
the sides of the triangle. 

If the angle opposite the side a be 60°, and if b } c be the 
remaining sides of the triangle, prove that 

8. Solve a triangle, having given the three sides. 
Given A = 36°, B = 72°, and a = 1 ; solve the triangle. 

9. The sides of a triangle are 2, 3, 4; find the least angle. 

* tan _1 .T= arc whose tangent is x. 



100 



TRIGONOMETRY. 



XI. 

(Cambridge, Eng. 2nd Previous Exam., Dec. 10, 1880. Time, 2\ hours.) 

1. Express in degrees, minutes, etc., (i.) the angle whose 
circular measure is -^tt; (ii.) the angle whose circular measure 
is 5. 

If the angle subtended at the centre of a circle by the side 
of a regular pentagon be the unit of angular measurement, by 
what number is a right angle represented ? 

2. Find, by geometrical constructions, the cosine of 45° and 
the sine of 120°. 

Prove that 

(sin 30° -f cos 30°) (sin 120° -f cos 120°) = sin 30°. 

3. If esc A = 9, find cot A and sec A. 

4. Prove that 

cos (180°+ A) = — eos A. 

Find the value of (i.) cot 840°; (ii.) sec37r. 

5. Assuming the formula for the sine of the sum of two 
angles in terms of the sines and cosines of the separate angles, 
find (i.) sin 75° ; (ii.) sin 3 A in terms of sin A. 

6. Prove the formulas : 

(1) cos 2 {A -B)- sin 2 (.4 + B) = cos 2 ^4 cos 2B, 

(2) 1 + tan# tan-§-# = sec#. 

7. Prove that 

cos A+ cos B = 2cos±(A + B) cos^A — B), 

-, cos 5 6 + cos 6 • i i 

and express — as a single term. 

cos 50 — cos 6 

8. Solve the equations : 

(1) 5 tan 2 .£ + sec 2 # = 7, 

(2) cos50 + cos30 = V2 cos40. 



EXAMINATION PAPERS. 101 

b 2 4- c 2 — a 2 

9. Prove that in any triangle cos A = — -E— 

2 be 

Obtain the formula for tan \ A in terms of the sides. 

10. Find an expression for the area of a triangle in terms 
of its sides. The lengths of the sides of a triangle are 3 feet, 
5 feet, and 6 feet ; what is its area ? 

11. Given that 

sin 38° 25' = 0.6213757, sin 38° 26' = 0.6216036 ; 
find the angle whose sine is 0.6215000. 



XII. 

{Cambridge, Eng. 2nd Previous Exam., Dec. 10, 1881. Time, 2\ hours.) 

1. Define the unit of circular measure. The ratio of the 
circumference of a circle to its diameter being 3.14159, find 
the circular measure of an angle of 126°. 

2. Define the tangent, cotangent, and cosecant of an angle. 
Find the tangent and cotangent of an angle whose cosecant is 
1.25. 

3. Trace the changes in sign and magnitude of sin A as A 
changes from 90° to 270°. 

4. Prove the following : 

(1) tan (it -f- A) = tan A, 

tan A — tan B 



(2) tea(A-£) = 



1 + tan A tan B 



(3) sin^ + sini? = eot ^-^ 
cos A — cos B 2 



102 



TRIGONOMETRY. 



5. Determine the value of cos 18°, and prove that 

cos 36° = cos 60° + cos 72°. 

6. Show that for certain values of the angles 



2 cos i A = Vl + sin A — V 1 — sin A. 

Is this formula true for values of A lying between 200° and 
220° ? and if not, how must it be modified ? 



7. Prove that in any triangle, with the usual notation, 
cos? A 



Is (s — a) 

and that the area is equal to 

Vs (s — a) (s — b) (s — 6). 
Show, also, that 

sin 2 A = cos 2 i? + cos 2 C-f 2 cos A cos B cos C. 

8. When one side of a triangle and the two adjacent angles 
are given, show how to solve the triangle. 

Find the greatest side of the triangle, of which one side is 
2183 feet, and the adjacent angles are 78° 14' and 71° 24'. 



SPHERICAL TRIGONOMETRY. 



CHAPTER V. 

THE RIGHT SPHERICAL TRIANGLE. 

§ 45. Introduction. 

The object of Spherical Trigonometry is to show how spheri- 
cal triangles are solved. To solve a spherical triangle is to 
compute any three of its parts when the other three parts are 
given. 

The sides of a spherical triangle are arcs of great circles. 
They are measured in degrees, minutes, and seconds, and 
therefore by the plane angles formed by radii of the sphere 
drawn to the vertices of the triangle. Hence, their measures 
are independent of the length of the radius, which may be 
assumed to have any convenient numerical value; as, for 
example, unity. 

The angles of the triangle are measured by the angles made 
by the planes of the sides. Each angle is also measured by 
the number of degrees in the arc of a great circle, described 
from the vertex of the angle as a pole, and included between 
its sides. 

The sides may have any values from 0° to 360° ; but in this 
work only sides that are less than 180° will be considered. 
The angles may have any values from 0° to 180°. 

If any two parts of a spherical triangle are either both less 
than 90° or both greater than 90°, they are said to be alike in 
kind; but if one part is less than 90°, and the other part 
greater than 90°, they are said to be unlike in kind. 



104 



TRIGONOMETRY. 



Spherical triangles are said to be isosceles, equilateral, equi- 
angular, right, and oblique, under the same conditions as 
plane triangles. A right spherical triangle, however, may- 
have one, two, or three right angles. 

When a spherical triangle has one or more_ of its sides equal 
to a quadrant, it is called a quadrantal triangle. 

It is shown in Solid Geometry, that in every spherical tri- 
angle 

I. The sum of the sides is less than 360°. 

II. The sum of the angles is greater than 180°. 

III. If, from the vertices as poles, arcs of great circles are 
described, another spherical triangle is formed so related to the 
first triangle that the sides of each triangle are supplements of 
the angles opposite to them in the other triangle. 

Two such triangles are said to be polar with, respect to each 
other. 

Let A, B, C (Fig. 35) denote the angles of one triangle ; a, b, c 

the sides opposite these angles 
respectively; and let A',B', C 
and a', b', c' denote the cor- 
responding sides and angles 
of the polar triangle. Then 
the above theorem gives the 
six following equations : 
A +a' = 180° 
B + b' = 180° 
G +V = 180° 
A' + a =180° 
B' + b = 180° 

a •+■ e = 180° 




Fig. 35. 



Exercise XIX. 

1. The angles of a triangle are 70 c 
sides of the polar triangle. 



80°, and 100°; find the 



THE RIGHT SPHERICAL TRIANGLE. 



105 



2. The sides of a triangle are 40°, 90°, and 125° ; find the 
angles of the polar triangle. 

3. Prove that the polar of a quadrantal triangle is a right 
triangle. 

4. Prove that, if a triangle have three right angles, the sides 
of the triangle are quadrants. 

5. Prove that, if a triangle have two right angles, the sides 
opposite these angles are quadrants, and the third angle is 
measured by the number of degrees in the opposite side. 

6. How can the sides of a spherical triangle be found in 
units of length, when the length of the radius of the sphere is 
known ? 

7. Find the lengths of the sides of the triangle in Example 
2, if the radius of the sphere is 4 feet. 



§ 46. Formulas Eelating to Eight Triangles. 

As is evident from § 45, Examples 4 and 5, the only kind of 
right triangle requiring further investigation is that which 
contains only one right angle. 

Let ABC (Fig. 36) be such a right triangle, and let A, B, C 
denote the angles of the tri- 
angle ; a, b, c, respectively, the 
opposite sides. 

Let be the right angle, and 
for the present suppose that 
each of the other parts is less 
than 90°. 

Let planes be passed through 
the sides, intersecting in the 
radii OA, OB, and 00; and 
for the sake of simplicity let 
the radius of the sphere be m „ ^ 

taken equal to 1. 

Also, let a plane perpendicular to OA be passed through 
B, cutting OA at E and 0(7 at D. Join BE, BE, and BE. 




106 



TRIGONOMETRY. 



B 



BE and DE are each ±OA (Geom. § 454) ; therefore ABED 

= A. The plane BDE is J_ the 
plane^O(7(Geom.§472); hence 
BD, which is the intersection 
of the planes BDE and BOC, is 
± the plane ^OC(Geom. § 475), 
therefore JL 00 and DE. Now 
cose = OE= OD X cos&, 
C and OD = cos a. 

.'. cose = cos a cos b. [37] 

sin a = BD = BE X sin A, 
and BE = sine. 



/ ' 1 \ 


V 


/ ' ' V 




/ / \l) 




^-^^^ ' '' 








\C- 




^^^ 


/ h 



Fig. 36 (&»*). 

Therefore, sin a = sin c sin A \ 

sin b = sin c sin B J 



[38] 



changing letters, sin 

Again, DE = BE X cos A, 

and also DE = OD X sin b ; 

hence, BE X cos A = OD X sin b ; 
that is, sin c cos A = cos a sin b ; 

whence cos J. = cos a sin b csce.* 

By substituting in this formula the value of cos a, obtained 
from [37], we obtain 

cos A = tan b cot c ) 
In like manner, cos B = tan a cot c ) L J 

And by substituting in the same formula the value of sin b, 
as given in [38], we obtain 

cos A = cos a sin B 
In like manner, cos B = cos b sin A 

Also, BD=DEx tan A, 

BD = sin a, 
DE = cos a sin b ; 
therefore, sin a = cos a sin b tan A 



[40] 



* To avoid fractions, cscc is written in place of its equal, 



THE RIGHT SPHERICAL TRIANGLE. 



107 



[41] 



whence, sin b = tan a cot A 

and, similarly, sin a = tan b cot B 

If in [37] we substitute for cos a and cos b their values from 
[40], we obtain 

cos c = cot A cot B. [42] 

In deducing these formulas, it has been assumed that all 
the parts of the triangle, except the right angle, are less than 
90°. But the formulas also hold true when this hypothesis is 
not fulfilled. 

Let one of the legs a be greater than 90°, and construct a 
figure for this case (Fig. 37) in the same manner as Fig. 36. 




Fig. 37. 



The auxiliary plane BBE will now cut both CO and AO 
produced beyond the centre ; and we have 

cos (180 -c) = OE=OBxcosb = cos (180 - a) cos b, 
a result which reduces to [37] if we substitute — cose in place 
of cos (180 — c) and —cos a in place of cos (180 — a). Like- 
wise, the other formulas, [38] -[42], hold true in this case. 

Again, suppose that both the legs a and b are greater than 
90°. In this case the plane BBE (Fig. 38) will cut CO pro- 
duced beyond O, and AO between A and O; and we have 
cose = OE=OBx cos (180 - b) 
== cos (180 - a) cos (180 - b) 
= cos a cos 6, 



108 TRIGONOMETRY. 



a result agreeing with [37], And the remaining formulas 
may be easily shown to hold true. 

Like results follow in all cases ; in other words, Formulas 
[37] -[42] are universally true. 

Exercise XX. 

1. Prove, by aid of Formula [37], that the hypotenuse of a 
right triangle is less than or greater than 90°, according as the 
two legs are alike or unlike in kind. 

2. Prove, by aid of Formula [40], that in a right spherical 
triangle each leg and the opposite angle are always alike in 
kind. 

3. "What inferences may be drawn respecting the values of 
the other parts : (i.) if c = 90° ; (ii.) if a = 90° ; (iii.) if c = 90° 
and a = 90° ; (iv.) if a = 90° and b = 90° ? 

Deduce from [37] -[42] the following formulas: 

4. tan 2 i b = tan ? (c — a) tan £ (c -j- a). 

5. tan 2 (45° - % A) = tan i (c - a) cot J (c + a). 

6. t&n 2 i B = sm (c — a)csc(c + a). 

7. tan 2 \ c = - cos (A + B) sec {A — B). 

8. tan 2 J a= tan [l(A + B)- 45°] tan [* (A~B) + 45°]. 

§47. Napier's Rules. 

The formulas deduced in § 46 express the relations between 
five parts of a right triangle, — the three sides and the two 
oblique angles. All these relations may be shown to follow 
from two very useful Pules, devised by Baron Napier, the 
inventor of Logarithms. 

For this purpose the right angle (not entering the formulas) 
is left out of account, and instead of the hypotenuse and the 
two oblique angles, their respective complements are employed ; 
so that the five parts considered by the Rules are : a, b, go. c, 
co. A, co. B. Any one of these parts may be called a middle 
part ; and then the two parts immediately adjacent are called 
adjacent parts, and the other two are called opposite parts. 



THE RIGHT SPHERICAL TRIANGLE. 



109 



Rule I. The sine of the middle part is equal to the product 
of the tangents of the adjacent parts. 

Rule II. The sine of the middle part is equal to the product 
of the cosines of the opposite parts. 

These Rules are easily remembered by the expressions, 
tan. ad. and cos. op. 

The correctness of these Rules may be shown by taking each 
of the five parts as middle 

b 



co.A 




co.B 



part, and comparing the 
resulting equations with the 
equations contained in For- 
mulas [37] -[42]. 

For example, let co. c be 
taken as middle part, then 
co. A and co. B are the ad- 
jacent parts, and a and b 
the opposite parts, — as is 
very plainly seen in Fig. 39, 
in which the order of the 
parts is shown by arranging them around the circumference 
of a circle. Then, by Napier's Rules : 

sin (co. c) = tan (co. A) tan (co. B), 
or cose = cot A cotB; 

sin(co. c) = cos a cos b, 
or cos c = cos a cos b ; 

results which agree with Formulas [37] and [42] respectively. 



Fig. 39. 



Exercise XXI. 

1. Show that Napier's Rules lead to the equations contained 
in Formulas [38], [39], [40], and [41]. 

2. What will Napier's Rules become, if we take as the five 
parts of the triangle, the hypotenuse, the two oblique angles, 
and the complements of the two legs ? 



110 TRIGONOMETRY. 



§ 48. Solution of Eight Triangles. 

By means of Formulas [37] -[42] we can solve a right tri- 
angle in all possible cases. In every case two parts besides 
the right angle must be given. 

Case I. Given the two. legs a and b. 

The solution is contained in Formulas [37] and [41]; viz: 

cos c = cos a cos b, 
tan A = tana esc b, 
tan B= tan b esc a. 

For example, let a = 27° 28' 36", b = 51° 12' 8" ; then the 
solution by logarithms is as follows : 

log cos a = 9.94802 
log cos 5 = 9.79697 



log cos c = 9.74499 

c = 56° 13' 40" 



logtana -=9.71604 
log esc b =0.10826 

log tan .4 = 9.82430 

.4 = 33° 42' 50" 



log tan b =10.09477 
log esc a = 0.33593 

log tan ^ = 10.43070 
B = 69° 38' 54" 



Case II. Given the hypotenuse c and the leg a. 
From Formulas [37], [38], and [39] we obtain 

cosft = cose? sec a, 
sin A = sin a esc c, 
cosi?== tana cote. 

Although two angles in general correspond to sin A, one 
acute the other obtuse, yet in this case the indetermination is 
removed by the fact that A and a must be alike in kind (see 
Exercise XX., Example 2). 



THE EIGHT SPHERICAL TRIANGLE. Ill 

Case III. Given the leg a and the opposite angle A. 
By means of Formulas [38], [40], and [41], we find, that 

sine = sin a esc A, 
sin b = tan a cot A, 
sin B = sec a cos A ; 

or, from [37] and [39], 

cos b = cose sec a, 
cos B — tan a cot c. 

When c has been computed, b and B are determined by 
these values of their cosines ; but, since c must be found from 
its sine, c may have in general two values which are supple- 
ments of one another. This case, therefore, really admits of 
two solutions. 

Case IV. Given the leg a and the adjacent angle B. 
Formulas [39], [40], and [41] give 

tan c = tan a sec B, 
tan b = sin a tan B, 
cos A = cos a smB. 

Case V. Given the hypotenuse c and the oblique angle A. 
From Formulas [38], [39], and [42] it follows that 

sin a = sin c sin A, 
tan b = tane cos A, 
cot B= cose tan A. 

Here a is determined by sin a, since a and A must be alike 
in kind (see Exercise XX., Example 2). 

Case VI. Given the two oblique angles A and B. 
By means of Formulas [40] and [42] we obtain 

cose == cot A cotB, 
cos« = cos A cscB, 
cos b = cosB cscA 



112 TRIGONOMETRY. 



Note 1. In Case I. {a and b given) the formula for computing c fails 
to give accurate results when c is very near 0° or 180° ; in this case it 
may be found with greater accuracy by first computing B, and then com- 
puting c, as in Case IV. 

Note 2. In Case II. (c and a given), if b is very near 0° or 180°, it 
may be computed more accurately by means of the derived formula 

tan 2 £ b = tan %(c + a) tan £ (c — a). 

And if A is so near 90° that it cannot be found accurately in the Tables 
it may be computed from the derived formula 

tan 2 (45° - £ A) = tan | (c - «) cot \ (c + a). 

In like manner, when B cannot be accurately found from its cosine we 
may make use of the formula 

tan 2 £ B = sin (c — a) esc (c + a). 

Note 3. In Case III. (a and A given), when the formulas for the 
required parts do not give accurate results, we may employ the derived 
formulas 

tan 2 (45° - J c) = tan %(A - a) cot %(A + a), 
tan 2 (45° - £ b) = sin (^i - a) esc (A + a) % 
tan 2 (45° - %B) = tan } (A - a) tan %(A + a). 

Note 4. In Case IV. (a and B given), if A is near 0° or 180°, it may 
be more accurately found by first computing b and then finding A. 

Note 5. In Case V. (c and A given), if a is near 90°, it may be found 
by first computing b, and then computing a by means of Formula [41]. 

Note 6. In Case VI. (A and B given), for unfavorable values of the 
sides greater accuracy may be obtained by means of the derived formulas 

tan 2 £c = - cos (A + B) sec (A - B), 

tan 2 1 a = tan [I {A + B) - 45°] tan [45° + \(A- B)\ 

tan 2 1 b = tan [\ (A + B) - 45°] tan [45° -\{A-B)\ 

Note 7. In Cases I., IV., and V., the solution is always possible; in 
Case II., in order that the solution should be possible, it is necessary 
and sufficient that sin a < sine; in Cases III. and VI., the conditions of 
possibility follow obviously from the equations employed ; in Case III., 
it is also necessary that a and A should be alike in kind. 



THE RIGHT SPHERICAL TRIANGLE. 



113 



Note 8. It is easy to trace analogies between the formulas for solving 
right spherical triangles and those for solving right plane triangles. The 
former, in fact, become identical with the latter if we suppose the radius 
of the sphere to be infinite in length ; in which case the cosines of the 
sides become each equal to 1, and the ratios of the sines of the sides and 
of the tangents of the sides must be taken as equal to the ratios of the 
sides themselves. 

If the formula required for any case is not remembered, it is 
always easy to find it by means of Napier's Eules. In apply- 
ing these Eules we must choose for the middle part that one 
of the three parts considered — the two given and the one 
required — which will make the other two either adjacent 
parts or opposite parts. 

For example : given a and B ; solve the triangle. 

First, represent the parts as in Fig. 40, and to prevent mis- 
takes mark each of the given 
parts with a cross. To find ° 

b, take a as the middle part ; 
then b and co. B are adja- 
cent parts; and by Rule I., 



sin a = tan 6 cot i?; 
whence, tan b = sin a tan B. 



co. A 




To find c, take co. B as 
middle part ; then a and 
co. c are adjacent parts ; and 
by Rule I., 

cos B = tan a cot c ; 
whence, tan c = tan a sec B 



To find A, take co. A as middle part ; then a and co. B are 
the opposite parts; and by Rule II., 

cos A = cos a sin B. 

In like manner, every case of a right spherical triangle may 
be solved. 



114 



TRIGONOMETRY. 



Exercise XXII. 
Solve the following right triangles, taking for the given 
parts in each case those printed in columns I. and II. : 



I. 


II. 


III. 


IV. 


V. 




a 


b 


c 


A B 


1 


36° 27' 


43° 32' 31" 


54° 20' 


46° 59' 43.2" 57°59'19.3" 


2 


86° 10' 


32° 40' 


87° 11' 39.8" 8S°11'57.S" 32°42'35.7" 


3 


50° 


36° 54' 49" 


59° 4' 25. 7" 63° 15' 13.1" ! 44°26'21.6" 


4 


120° 10' 


150° 59' 44" 


63° 55' 43.2" 


105° 44' 21.25" 147°19 / 47.14" 




c 


a 


b 


A 


B 


-5 


55° 9' 32" 


22° 15' 7" 


51° 53' 


27° 28' 25.7" 


73° 27' 11.16" 


6 


23° 49' 51" 


14° 16' 35" 


19° 17' 


37° 36' 49.4" 


54° 49> 23.3" 


7 


44° 33' 17" 


32° 9' 17" 


32° 41' 


49° 20 7 16.4" 


50° 19' 16" 


8 


97° 13' 4" 


132° 14' 12" 


79° 13' 38.2" 


131° 43' 50" 


81° 5S' 53.3" 




a 


A 


c 


b 


B 


9 


77° 21' 50" 


83° 56' 40" 


78° 53' 20" 


28° 14' 31.3" 


28° 49' 57.4" 








10P 6' 40" 


151° 45' 28.7" 


151° 10' 2.6" 


10 


77° 21' 50" 


40° 40' 40" 


impossible ; 


why? 






a 


* 


c 


b 


A 


11 


92° 47' 32" 


50° 2' 1" 


91° 47' 40" 


50° 


92° 8' 23" 


12 


2° 0'55" 


12°4(y 


2° 3' 55.7" 


0° 27' 10.2" 


77° 20' 25.4" 


13 


20° 20' 20" 


38° 10' 10" 


25° 14' 38.2" 


15° 16' 50.4" 


54° 35' 16.7" 


14 


54° 30' 


35° 30' 


59° 51' 20.8" 


30° 8' 39.2" 


70° 17' 35" 




c 


A 


a 


b 


B 


15 


69° 25' 11" 


54° 54' 42" 


50° 


56° 50' 49.3" 


63° 25' 4" 


16 


112° 48' 


56° 11' 56" 


50° 


127° 4' 30" 


120° 3' 50" 


17 


46° 40' 12" 


37° 46' 9" 


26° 27' 24" 


39° 57' 41.5" 


62° 0' 4" 


18 


118° 40' 1" 


128° 0' 4" 


136° 15' 32.3" 


4S° 23' 38.4" 


:-27' 4.3" 


A 


B 


a 


b 


c 


19 


63° 15' 12" 


135° 33' 39" 


49° 59' 56" 


143° 5' 12" 


120° o3 34.3" 


20 


116° 13' 12" 


116° 31' 25" 


120° 10' 3" 


119° 59' 46" 


75° 26' 58" 


21 


16° 59' 42" 


57° 59' 17" 


36° 27' 


43° 32' 30". 


54° 20» 


22 


90° 


88° 24' 35" 


90° 


55° 24' 35" 


90° 



THE EIGHT SPHERICAL TRIANGLE. 115 

23. Define a quadrantal triangle, and show how its solution 
may be reduced to that of the right triangle. 

24. Solve the quadrantal triangle whose sides are : 

a = 174° 12' 49.1", 6 = 94°8'20", e = 90°. 

25. Solve the quadrantal triangle in which 

£ = 90°, A = 110° 47' 50", .5 = 135° 35' 34.5". 

26. Given in a spherical triangle A, C, and c = 90° ; solve 
the triangle. s 

27. Given ^4 = 60°, (7=90°, and c = 90°; solve the tri- 
angle. 

28. Given in a right spherical triangle, A = 42° 24' 9", 
^ = 9°4'11"; solve the triangle. 

29. In a right triangle, given a = 119° 11', B = 126° 54'; 
solve the triangle. 

30. In a right triangle, given c = 50°, 6 = 44° 18' 39"; solve 
the triangle. 

31. In a right triangle, given J=156° 20' 30", a=65°15'45"; 
solve the triangle. 

32. If the legs a and b of a right spherical triangle are 
equal, prove that cos a = cot A = Vcos<?. 

33. In a right triangle prove that cos 2 A X sin 2 c = sin (c— a) 

34. In a right triangle prove that tana cose = sin b cot B. 

35. In a right triangle prove that 

sin 2 A = cos 2 i? -f sin 2 a sin 2 i?. 

36. In a right triangle prove that 

sin (b -f c) — 2 cos 2 i A cos b sin c. 

37. In a right triangle prove that 

sin (c — b) = 2 sin 2 1 A cos b sin c. 

38. If, in a right triangle, p denote the arc of the great circle 
passing through the vertex of the right angle and perpendic- 
ular to the hypotenuse, m and n the segments of the hypote- 
nuse made by this arc adjacent to the legs a and b respectively, 
prove that (i.) tan 2 a = tanctanra, (ii.) sin 2 ^> = tanmtanw. 



116 TRIGONOMETRY. 



§49. Solution of the Isosceles Spherical Triangle. 

If an arc of a great circle is passed through the vertex of 
an isosceles spherical triangle and the middle point of its base, 
the triangle will be divided into two symmetrical right tri- 
angles. In this way the solution of an isosceles spherical 
triangle may be reduced to that of a right spherical triangle. 

In a similar manner the solution of a regular spherical 
polygon may be reduced to that of a right spherical triangle. 
Arcs of great circles, passed through the centre of the polygon 
and its vertices, divide it into a series of equal isosceles tri- 
angles ; and each one of these may be divided into two equal 
right triangles. 

Exercise XXIII. 

1. In an isosceles spherical triangle, given the base b and 
the side a; find A the angle at the base, B the angle at the 
vertex, and h the altitude. 

2. In an equilateral spherical triangle, given the side a; 
find the angle A. 

3. Given the side a of a regular spherical polygon of n 
sides ; find the angle A of the polygon, the distance R from 
the centre of the polygon to one of its vertices, and the dis- 
tance r from the centre to the middle point of one of its sides. 

4. Compute the dihedral angles made by the faces of the 
five regular polyhedrons. 

5. A spherical square is a spherical quadrilateral which 
has equal sides and equal angles. Its two diagonals divide it 
into four equal right triangles. Find the angle A of the 
square, having given the side a. 



CHAPTEE VI. 



THE OBLIQUE SPHERICAL TRIANGLE. 

§ 51. Fundamental Formulas. 

Let ABC (Fig. 41) be an oblique spherical triangle, a, b } c 
its three sides, A,B,C the angles 
opposite to them, respectively. 

Through draw an arc CD of 
a great circle, perpendicular to 
the side AB, meeting AB at D. 
For brevity let CD=p, AD=m, 
BD = n, ZACD = x, /.BCD 

1. By §46 [38], in the right 
triangles BDC and ADC, 

sinp = sin a sin B, 
and sinp = sin b sin A. 

Therefore, sin a sinB = sinb sin A 

similarly, sin a sinO = sine sin A 

and sinb sinO = sine sinB 




[43] 



These equations may of course also be written in the form 
of proportions ; as, for example, 

sin a : sin b = sin A : sin B. 

In Fig. 41 the arc of the great circle CD cuts the side AB 
within the triangle. In case it cut AB produced without the 
triangle, sin (180° - A\ sin (180° - B), or sin (180° - C), would 
be employed in the above proof instead of sin^, sini?, or 
sin C These sines, however, are equal to sin A, sin B, and 
sin C, respectively, so that the Formulas [43] hold true in all 
cases. 



118 TRIGONOMETRY. 



2. In the right triangle BBC, by §46 [37], 

cos a = cosp cosw = cosp cos(c — ra), 
or (§ 32) cos a = cosp cose cosm -f- cosp sine sinm. 

Now, cosp cosm = cos 6; (§46 [37]) 

whence, cosp = cos b sec m, 

and cosp sinm = cos b tanra 

= cos& tan 6 cos J. (§46 [39]) 

= sin5 cos A. 

Substituting these values of cosp cosm and cosp sinm in 
the value of cos a, we obtain 

cosa = cosb cose + sinb sine cos A ] 
and similarly, cosb = cosa cose + sin a sine cos B > • [44] 

cose = cosa cosb -f- sina sinb cosO J 

3. In the right triangle ADC, by §40 [41], 

cos A = cosp sin x = cosp sin ( C— y), 
or (§ 32) cos A = cosp sin cosy — cosp cos C siny. 

Now, cosp sin y = cos B; (§ 46 [40]) 

whence, cosp = cos B esc y, 

and cosp cosy = cosi? coty 

= cos B tan B cos a (§ 46 [42]) 

= sini? cosa. 

Substituting these values of cosp siny and cosp cosy in the 
value of cos A, we obtain 

cosA = — cosB cosO+sinB sinO cosa^l 

and similarly, cosB == — cos A cosO -f sin A sinO cosb \ • [45] 
cosO = — cos A cosB -j-sinA sinB cose J 

Formulas [44] and [45] are also universally true ; for the 
same equations are obtained when the arc CD cuts the side 
AB without the triangle. 



THE OBLIQUE SPHERICAL TRIANGLE. 119 



Exercise XXIV. 

1. What do Formulas [43] become if A = 90°? if B = 90°? 
if 0= 90°? ifa = 90°? if A = £=90°? ifa = e- = 90°? 

2. What does the first of [44] become if A = 0° ? if A = 90° ? 
if ^4 = 180°? 

3. From Formulas [44] deduce Formulas [45], by means of 
the relations between polar triangles (§ 45). 



§ 52. Formulas for the Half Angles and Sides. 
From the first equation of [44], 

cos a — cos b cose 



whence, 



cos A 



1 — cos A 



sin b sine 
sin b sin e -j- cos b cos e — cos a 



sin b sine 
_ cos (b — e) — cos a 
sin 6 sine 

-, , A sin & sine — cos 5 cose -f- cos a 

1 + cos A = : — - — : ! 

sin b sin e 

_ cos a— cos(&- + <?) . 

sine- sine 

or, by §34 [16] and [17], and § 35 [23], 

sin 2 1 A = sin I (a-\-b — c) sin £ (a — b -f- e) esc e- esc e, 
cos 2 2 -^4 = sin ? (a+ & + e) sin j> : (£- -f- c— r?) esc b esc e. 

Now, let %(a-\-b + c) = s; 

whence, % (b -j- c — a) = s — a, 

£ (a — b -J- c) — 5 — 5, 

J (a + & — c) = s — e. 



120 



TRIGONOMETRY. 



Then, by substitution and extraction of the square root, 

[46] 



sin £ A = Vsin(s — b) sin(s — c) cscb cscc 

cos £ A= Vsins sin(s — a) cscb cscc 

tan J A= Vcsc s esc (s — a) sin(s — b) sin(s — c) 

In like manner it may be shown that 

sin £B = Vsin(s — a) sin(s — c) csca cscc, 

cos £B = Vsins sin(s — b) csca cscc, 

tan i B = Vcscs esc (s — b) sin (s — a) sin (s — c). 

sin £0 = Vsin (s — a) sin(s — b) csca cscb, 

cos i = Vsins sin(s — c) csca cscb, 

tan J = Vcsc s esc (s — c) sin(s — a) sin(s — b). 

Again, from the first equation of [45], 



cosa : 
whence, 

1 — cosa = 



cosi? cos (7-f cos A . 
sin B sin C 

sin B sin C— cos B cos Q — cos^i 



1 + cos a 



sin B sin C 

sin B sin C~\- cos B cos C j- cos A 
sin B sin C 



If we place £ ( A + B -f C) = S, and proceed in the same 
manner as before, we obtain the following results : 



sin£a =V— cosS cos(S— A) cscB cscO 
cosi a =Vcos(S — B) cos(S — 0) cscB cscO 



tania ^V^cosS cos(S— A) sec(S — B) sec(S — 0) 



\.[i7] 



THE OBLIQUE SPHERICAL TRIANGLE. 121 

And, in like manner, 

sin £b = V— cosS cos(S — B) esc A csoO, 
cos?b=Vcos(S — A) cos(S — 0) cscAcscO, 



A, 



tan £ b = V -cos S cos (S - B) sec (S - A) sec (S - 0). 

sin Jc = V— cos S cos (S — 0) esc A cscB, 
cos£c =Vcos(S— A) cos(S — B) esc A esc B, 
tanic = V —cos S cos (S — 0) sec (S — A) sec (S — B). 

§ 53. Gauss's Equations and Napier's Analogies. 

By §31 [5], 

cos^(A-{-B) = cos-|- A cos^B — sin£ A sin-|-i?; 

or, by substituting for cos \A, cos^B, sin \A, $m\B y their 
values given in § 52, and reducing, 



cos 



i ( a 4- Tft — H ns sin(s— a) J sins sin(s— b) 

\ sin b sin c \ sin a sin c 

[ sin (s — b) sin (s — c) It 

\ sin b sine? \ 



(s— 5)sin(s — c) v ^ j sin (5 — a) sin (5 — c) 
sin a sin c 



sing— sin (s—c) J sin (s — a) sin (s — b) 

sin <? \ sin a sin 6 

This value, by applying §§ 33 [12], 35 [21], and observing 
that the quantity under the radical is equal to sin I C } becomes 

1 / a i t>\ 2 sin \c cos (s — 4-c) . ,^ 
cos%(A + B) = — —2— — v 2 / S in I (7, 
2sm}c cos-Jc 

which, by cancelling common factors, multiplying by cos-^-c, 
and observing that s — ^c = ^(a-\-b), reduces to the form 

cos^(A + B) cos^c = cos ^ (a -\-b) sin ^ C. 



122 



TRIGONOMETRY. 



By proceeding in like manner with the values of 

smi(A + £), cos$(A-B), and sin£(4 - B), 

three analogous equations are obtained. 

The four equations, 

cos £ (A + B) cos \ c == cos i (a + b) sin } 
sin 5 (A + B) cos I c = cos \ (a — b) cos \ 
cos \ (A — B) sin \ c = sin \ (a -f b) sin \ 
sin J (A — B) sin \ c = sin £ (a — b) cos \ 



[48] 



are called Gauss's Equations. 

By dividing the second of Gauss's Equations by the first, 
the fourth by the third, the third by the first, and the fourth 
by the second, we obtain 

co S i(a-b) coti(r 

cos 2 (a + b) 



tan*(A + B) 



tanJ(A-B) 
tan \ (a -j- b) 
tan \ (a — b) 



_ sin \ (a — b) 

sin \ (a -f- b) 

__cosj(A-B) 



cot JO 



tanic 



cosi(A+B) 

sk ^ A - B )tanjc 
sini(A+B) * 



[49] 



These equations are called Napier's Analogies. 

In the first equation the factors cos i (a — b) and cot \ C are 
always positive; therefore, tan ^ {A -f B) and cos -J- (a -\-b) 
must always have like signs. Hence, if a + b < 180°, and 
therefore cos h (a + 5) > 0, then, also, tan £ (-4 -{- i?) > 0, and 
therefore J. -f ^ < 180°. Similarly, it follows that if 
« + &>180 o , then, also, ^ + -£>i80°. If a + 6 = 180°, 
and therefore cos £ (a -f b) = 0, then tan J (^4 -f 5) = oo ; 
whence i(A + £) = 90°, and A + B= 180°. 

Conversely, it may be shown from the third equation, that 
a + b is less than, greater than, or equal to 180°, according as 
A -f- B is less than, greater than, or equal to 180°. 



THE OBLIQUE SPHERICAL TRIANGLE. 



m 



§ 54. Case I. 

Oiven two sides, a and b, and the included angle C. 
The angles A and B may be found by the first two of 
Napier's Analogies ; viz. : 

CQB*(a-*) cotH 7. 



tan \\Jl -i-B) 
tun I (A -B) 



cos h (a -f- b) 
_sinKa-6) cot|a 



sin i (a -J- b) 

After A and B have been found, the side c may be found 
by [43] ; but it is better to use for this purpose Gauss's Equa- 
tions, because they involve functions of the same angles that 
occur in working Napier's Analogies. Any one of the equa- 
tions may be used ; for example, from the first we have 

! cos \{a -f- b) • , n 

cos$c = ) ( sin \ C 

cos h (A+B) 

Example. a = 73° 58' 54", 

b = 38° 45' 0", 

C= 46° 33' 41", 
log cos £ (a -b) =9.97914 
logsec|(a + 6) =0.25658 
log cot \C =0.36626 

log tan $ (4 + B) = 0.60198 
log sec $(A+B) = 0.61547 
logcosJ(a + 6) =9.74342 
logsin^C =9.59686 



log cos \ c 



= 9.95575 
= 25° 25' 40" 



eref 


ore, §(a 


-6) = 


= 17° 36' 57" 




$(a + b) = 


= 56° 21' 57" 




ho 




= 23° 16' 50.5" 


log 


sin ${a — 


&) = 


= 9.48092 


log 


CSC 2 (« + 


6) - 


= 0.07956 


log 


cot $C 




= 0.36626 


log 


tan £ (A - 


-£) = 


= 9.92674 




H^ + ^)- 


= 75° 57' 40.7" 




1(4- 


-£) = 


= 40° 11' 25.6" 






^4 = 


= 116° 9' 6.3" 






5 = 


= 35° 46' 15.1" 






c = 


= 50° 51' 20" 



If the side c only is desired, it may be found from [44], 
without previously computing A and B. But the Formulas 
[44] are not adapted to logarithmic work. Instead of chang- 
ing them to forms suitable for logarithms, we may use the fol- 
lowing method, which leads to the same results, and has the 
advantage that, in applying it, nothing has to be remembered 
except Napier's Eules : 



124 



TRIGONOMETRY. 




Make the triangle (Fig. 42), as in § 51, equal to the sum 

(or the difference) of two right 
triangles. For this purpose, 
through B (or A, but not C) 
draw an arc of a great circle 
perpendicular to AC, cutting 
AC at I). Let BD=p,CD=m, 
AD = n; and mark with crosses 
the given parts. 
By Rule I., 

cos C = tan m cot a, 
whence tanm = tana cos C. 
By Rule II., 
cos a = cos m cosp, whence cos p = cos a sec m. 
cos c = cos n cos p, whence cos^ = cos c sec n. 
cos c sec n = cos a sec ra ; 
n = b — m, cos c = cos a sec m cos (5 — ?n). 

It is evident that c may be computed, with the aid of loga- 
rithms, from the two equations 
tan m = tan a cos C, 
cose = cosa seem cos (6 — m). 

Example. Given a = 97° 30' 20", b = 55° 12' 10", C= 39° 
58'; findc. 

log tan a = 0.88025 (n) 
log cos O =9.88447 



Therefore 
or, since 



log tan ra = 0. 76472 (n) 
m = 99° 45' 14" 
&-m = -44°33' 4" 



log cos a = 9.11602 (n) 
logcos(6-ra) = 9.85286 
logsecm = 0.77103 (rc) 
logcos c = 9.73991 

c = 56° 40' 20" 



Exercise XXV. 

1. Write formulas for finding, by Napier's Rules, the side 
a when b, c, and A are given, and for finding the side b when 
a, c, and Z? are given. 



THE OBLIQUE SPHERICAL TRIANGLE. 



125 



2. Given a = 88° 12' 20", b =■ 124° 7' 17", C= 50° 2M" 
find J. = 63° 15' 11", B = 132° 17' 59", c = 59° 4' 18". 

3. Given a = 120° 55' 35", b --= 88° 12' 20", 0= 47° 42' 1" 
find A = 129° 58' 3", B '=. 63° 15' 9", c = 55° 52' 40". 

4! Given 5 -63° 15' 12", c = 47°42'l", A = 59° 4' 25" 
find 5 = 88° 12' 24", C= 55° 52' 42", a = 50° 1' 40". 

5. Given b = 69° 25' 11", c = 109° 46' 19", ^-54° 54' 42" 
find £ = 56° 11' 57", (7= 123° 21' 12", a = 67° 13'. 



§ 55. Case II. 

(xwen two angles, A and B, and the included side c. 
The sides a and b may be found by the third or fourth of 
Napier's Analogies, 

cos J (^4 + B) 

tanK«-S) = SB-t-£4xf fa**'."' 

sin 5 ( J. -f i?) 

and then the angle C may be found by one of Gauss's equa 
tions ; as, for instance, the second, which gives 

&mh(A + B) 



cos i(a- 


-b)- 


2 V. 




Example. .4 = 107° 47' 7" 
B= 38° 58' 27" 
c= 51° 41' 14" 
log cos £(.!-£) = 9.91648 
logsecJ(4 + J5) = 0.54359 
log tan \c =9.68517 


log sin 
log esc 
log tar 


HAS)- 

i(A + B) = 

\c = 

i(A-B) = 

lie 


= 34° 24' 20" 

= 73° 22' 47" 
= 25° 50' 3 7" 

= 9.75208 
= 0.01854 
= 9.68517 


logtanJ(a + 6) =0.14524 


log tar 


^{d-b) = 


= 9.45579 


log sin J (.4 + B) = 9.98146 
log sec \ (a -b) =0.01703 
log cos %c =9.95423 




J(a-6) = 
fa- 


= 54° 24' 24.4" 
= 15° 56' 25.6" 
= 70° 20' 50" 


log cos \C =9.95272 

£C=26°14'52.5" 




£ 


= 38° 27' 59" 
= 52° 29' 45" 



126 



TRIGONOMETRY. 



If the angle C alone is wanted, the best way is to decompose 
the triangle into two right triangles, and then apply Napier's 
Rules, as in Case I., when the side c alone is desired. 

Let(Fig. ±3) Z ABB = x,ZCBD = 7/, BD = p; then, 

By Rule I., 

cos<? = cot x cot A, 
whence cota? = tan^4 cose. 

By Rule II., 

cos .4 = cosp sin x, 
whence cos p = cos A esc x. 

cos C = cosp siny, 
whence cos p = cos Ccscy. 

Therefore, 

cos = cos A csc.t siny = cos A csc.-e sin(i? — x). 
It is clear that O may be computed from the equations 
cot a; = tan A cos c, 
cosC=cosA esc x sin (B — x). 

Example. Given A = 35° 46' 15", 5 = 115° 9' 7", c = 5i°2'; 
find C. 




log tan .4 = 9.85760 
log cos c =9.79856 
logcot# =9.65616 

x =65° 37' 35" 
B~x =49° 31' 32" 


log cos A =9.90992 

log sin (£-*) = 9.88122 

log csc # = 0.04055 


log cos C= 9.83099 

0= 47° 20' 30" 



Exercise XXVI. 

1. What are the formulas for computing A when B, C, and 
a are given; and for computing B when A, C, and b are 
given ? 

2. Given A = 26° 58' 46", B = 39° 45' 10", c = 154° 46' 48" ; 
find a -37° 14' 10", b = 121° 28' 10", (7= 161° 22' 11". 



THE OBLIQUE SPHERICAL TRIANGLE. 127 

3. Given .4 = 128° 41' 49", B= 107° 33' 20", c=124° 12' 
' 31"; find a= 125° 41' 44", b -82° 47' 34", 0= 127° 22'. 

^4. Given 5 = 153° 17' 6", C= 78° 43' 36", a =--86° 15'15"; 
"find 6 = 152° 43' 51", c = 88° 12' 21", A = 78° 15' 48". 

5. Given ^ = 125° 41' 44", (7= 82° 47' 35", 5 = 52° 37' 57"; 
find a= 128° 31' 46", c = 107° 33' 20", 5 =55° 47' 40". 



Q. 

§ 56. Case III. U 



Given two sides a and b, and the angle A opposite to one of 
them. 

The angle B is found from [43], whence we have 

sin B = sin A sin b esc a. 

When B has been found, and c may be found from the 
fourth and the second of Napier's Analogies, from which we 
obtain 

tan \ c — Rm \) J. tan £ (a — b), 

sin h {A — B) 

sin f (a — o) 

The third and the first of Napier's Analogies may also be 
used for the same purpose. 

Note 1. Since B is determined from its sine, the problem in general 
lias two solutions ; and, moreover, in case sin B> 1, the problem is impos- 
sible. By geometric construction it may be shown, as in the correspond- 
ing case in Plane Trigonometry, under what conditions the problem 
really has two solutions, one solution, and no solution. But in practical 
applications a general knowledge of the shape of the triangle is known 
beforehand ; so that it is easy to see, without special investigation, which 
solution (if any) corresponds to the circumstances of the question. 

It can be shown that there are two solutions, when 
^L<90°, a + J<180°, anda<Z>, 
or, when ^4>90°, a + J>180°, and a > b. 



128 



TRIGONGMETRY. 



Note 2. The side c or the angle C may be computed, without first 
finding B, by means of the formulas 

tan m = cos A tan b, and cos (c — ra) = cos a X sec b X cos m, 

cot a; = tan A cos 6, and cos(C— x) = cot a X tan& X cos#. 

These formulas may be obtained by resolution of the triangle into 

right triangles, and applying Napier's Rules ; m is equal to that part of 

the side c included between the foot of the perpendicular from C and the 

vertex A, and x is equal to the corresponding portion of the angle 0. 

Example. Given a = 57° 36' b = 31° 12', A = 104° 25' 30". 



In this case A> 90°, 

and a + 6<180°; 

therefore, A + B < 180° ; 

hence, B< 90°, 

and only one solution. 

a + b = 88°50' 
a- b =26° 26' 
A + B = 140° 51' 53" 
A-B= 67° 59' 7" 
logsin|(^ + -5) = 9.97416 
log esc £ (A -£) =0.25252 
logtan|(a-6) =9.37080 
logtan£e = 9.59748 

%c = 21° 35' 38" 
c = 43° 11' 16" 



log sin A = 9.98609 
log sin b =9.71435 
log esc a =0.07349 

log sin B = 9.77393 

B = 36° 27' 20" 

\ (a + b) = 44° 25' 

i(a - b) = 13° 13' 

%(A+ B) = 70° 26' 25" 

£ (A -5) = 33° 59' 5" 

log sin $(a + b) =9.84502 

logcsc£(a-6) =0.64086 

log tan | (A -£) = 9.82873 

log cot|C= 0.31461 

| C =25° 51' 15" 
C =51° 42' 30" 



Exercise XXVII. 

1. Given a = 73° 49' 38", b = 120° 53' 35", A = 88° 52' 42"; 
find _£= 116° 42' 30", c = 120° 57' 27", tf= 116° 47' 4". 

2. Given a =150° 57' 5", 5 = 134° 15' 54", ^4 = 144° 22' 42"; 
find jBx = 120° 47' 45", c x = 55° 42' 8", C, = 97° 42' 55.4" ; 



B 9 = 59° 12' 15", 



23° 57' 17.4", C 2 = 29° 8' 39". 



, 3. Given a'= 79° 0' 54.5", b = 82° 17' 4", A = 82° 9' 25.8" ; 
find B = 90°, c = 45° 12' 19", C= 45° 44'. 

4. Given a = 30° 52' 36.6", J = 31° 9' 16", J. = 87° 34' 12" ; 
show that the triangle is impossible. 



THE OBLIQUE SPHERICAL TRIANGLE. 129 



§ 57. Case IV. 

Given two angles A and B, and the side a opposite to one of 
them. 

The side b is found from [43], whence 

sin b = sin a sin B esc A. 

The values of c and C may then be found by means of Na- 
pier's Analogies, the fourth and second of which give 

taa J, =!iaMA±|) tan i (a - b), 

cotiC= 3in f( a + ^ tani(^--g). 
sin \ (a — o) 

Note 1. In this case, also, an unknown part is found from its sine ; 
and it may be shown that, under certain conditions, the problem is im- 
possible, or that it admits of two solutions. In practice, the ambiguity is 
usually removed by the circumstances of the question. If sin6>l the 
problem of course is impossible ; and it may be shown that there are two 
solutions, when 

a<90°, A+B<IW°, and^.<5, 
or a>90°, ^ + 5>180°, and A>B. 

Note 2. By proceeding as indicated in Case III., Note 2, formulas for 
computing c or 0, independent of the side b, may be found ; viz. : 

tan m = tan a cos B, and sin (c — m) = cot A tan B sin w, 
cot x = cos a tan B, and sin (0— x) = cos A sec B sin x. 

In these formulas m = BD, x~= Z BCD, D being the foot of the per- 
pendicular from the vertex C. 



Exercise XXVIII. 

1. Given A = 110° 10', .5 = 133° 18', a =147° 5' 32"; find 
b = 155° 5' 18", c = 33° l r 36", C= 70° 20' 40". 

2. Given ^ = 113°39'21", .£=123° 40' 18", «=65°39'46"; 
find b = 124° 7' 20", c = 159° 53' 2", C= 159° 43' 35". 



130 



TRIGONOMETRY. 



3. Given JL=100°2'11.3 H , £=98°30'28", a = 95°20'38.7"; 
find b = 90°, c = 147°41'43", C=148°5'33". 

4. Given A = 24° 33' 9", B = 38° 0' 12", a = 65° 20' 13" ; 
show that the triangle is impossible. 



§ 58. Case V. 

Given the three sides, a, b, and c. 

The angles are computed by means of Formulas [46], and 
the corresponding formulas for the angles B and C. 

The formulas for the tangent are in general to be preferred. 
If we multiply the equation 

tan £ A = Vcsc s esc (s — a) sin (s — b) sin (s — c) 



by the equation 
and put 



sin (s — a) 

» 

sin (s — a) 



Vcsc s sin (s — a) sin (s — b) sin (s — c) = tan ?% 

and also make analogous changes in the equations for tan h B 
and tan£C, we obtain 

tan %A = tan r esc (s — a), 
tan £ B = tan r esc (s — 5), 
tan J C — tanr esc (s — c), 

which are the most convenient formulas to employ when all 
three angles have to be computed. 



log esc s= 0.00448 

log esc (s- a) = 0.28978. 

logsin(s-Z>)= 9.84171 

logsin(s-c)= 9.08072 



Example 


1. a = 50° 54' 32" 




b= 37° 47' 18" 




c= 74° 51' 50" 




2s = 163° 33' 40" 




s= 81° 46' 50" 




s -a= 30° 52' 18" 




S _Z> = 43° 59' 32" 




a _c = 6° 55' 0" 



2)19.21669 

log tan M= 9-60835 
1^ = 22° 5' 20" 
^l = 44 o 10 / 40" 



THE OBLIQUE SPHERICAL TRIANGLE. 



13) 



Example 2. 



a = 124° 12' 31" 
6= 54° 18' 16" 

c= 97°12 / 25 // 

27 



s = 137° 51' 36" 
s-a= 13° 3^ 5" 
s-Z> = 83° 33' 20" 
s-c= 40° 39' 11" 



log tan \A = 0.30577 
log tan ^ = 9.68145 
logtanJC = 9.86480 



u= 


63° 41' 3.8" 


*5= 


25° 39' 5.6" 


*0= 


36° 13' 20.1" 



275° 43' 12" 

log sin (s- a) = 9.37293 

log sin (s- b) = 9.99725 

logsin(s- c) = 9.81390 

logcsc s = 0.17331 

logtanV = 9.35739 

log tan r = 9.67870 

~^=127°22' 7" 
£ = 51° 18' 11" 
C= 72° 26' 40" 

Exercise XXIX. 

1. Given a = 120° 55' 35", b = 59° 4' 25", <? = 106° 10' 22" 
find J. = 116° 44' 49", ^ = 63° 15' 14", 0= 91° 7' 21". 

2. Given a -50° 12' 4", 6 = 116° 44' 48", c = 129° 11' 42" 



find ^4 = 59° 4' 28", ^ 



12", (7=-- 120° 4' 52". 



3. Given a = 131° 35' 4", Z> =-- 108° 30' 14", c= 84° 46' 34" 
find ^ = 132° 14' 21", £ = 110° 10' 40", 0= 99° 42' 24". 

4. Given a = 20° 16' 38", b = 56° 19' 40", c = 66° 20' 44" 
find 4 = 20° 9' 54", .£=55° 52' 31", C=- 114° 20' 17". 



§59. Case VI. 

Given the three angles, A, B, and C. 

The sides are computed by means of Formulas [47], and 
the corresponding formulas for the angles B and C. The 
formulas for the tangents are in general to be preferred. 

If we multiply the equation 

tan£a= V— cos 8 cos (8 — A) sec (8— B)sec(8—C) 



by the equation 
and put 



secQS y — A) 
Bee (8 -A) 



y/-cos8sec(8-A)Bec(8- B) sec (8 - C) = tan IZ t 



132 



TEJGONOMETRY. 



and also make analogous changes in the equations for tan J b 
and tan £ c, we obtain 

tan J a = tan R cos (S — A), 

tan %b = tan i2 cos (S — JB) t 

tan £ c = tan i? cos (8 — C), 
which are the most convenient formulas to use in case all 
three angles have to be computed. 



Example I. A = 220° 


log cos #=9.53405 (n) 


B = 130° 


logcos(£-4) = 9.93753 


0= 150° 


log sec (S-B) = 0.30103 (n) 


28=500° 


log sec^S- C) = 0.76033 (n) 


S=250° 


2)0.53294 


S-A = 30° 


logtan|a = 0.26647 


£-£ = 120° 


£a = 61° 34' 6" 


£-C = 100° 


a =123° 8' 12" 



Note. Here the effect, as regards algebraic sign, of three negative 
factors, is cancelled by the negative sign belonging to the whole fraction. 

Example II. A= 20° & 56" 
B = 55°52'32" 
C = 114°20'14" 



2 #=190° 22' 42" 

log cos #=8.95638 (n) 

log sec (£-^) = 0.58768 

log sec (#-5) = 0.11143 

log s ec(#-C) = 0.02472 

log tan 2 i2 = 9.68021 

log tan R = 9.84010 



S= 95° 11' 21" 
8-A= 75° 1'25" 
S-B= 39° 18' 49" 
S-C = -19° 8' 53" 



logtanJa = 9.25242 
logtan£Z> = 9.72867 
log tan \c = 9.81538 



\d = 


= 10° 


8' 18.9" 


*& = 


= 28° 


9' 50.4" 


£c = 


= 33° 


10 / 21.3" 



a = 20° 16' 38" 
6 = 56°19 / 41" 
c=66°2(y43" 



Exercise XXX. 
.5 = 110°, <7=80 c 



1. Given ^4 = 130° 
find a = 139° 21' 22", b = 126° 57' 52", c = 56°51'48". 

2. Given J. = 59° 55' 10", B = 85° 36' 50", O = 59° 55' 10" 
find a = 128° 42' 29", 6 = 64° 2' 47", c = 1 28° 42' 29". 



THE OBLIQUE SPHERICAL TRIANGLE. 133 



3. Given A = 102° 14' 12", B = 54° 32' 24", C= 89° 5' 4G" ; 
find a = 104° 25' 9", b = 53° 49' 25", c = 97° 44' 24". 

4. Given ^4 = 4° 23' 35", B = 8° 28' 20", (7= 172° 17' 56" ; 
find a = 31° 9' 11", 6 = 84° 18' 23", c= 115° 9' 56". 

§60. Area op a Spherical Triangle. 

I. When the three angles A, B, C, are given. 

Let R = radius of sphere, 

E= the spherical excess = A + B-\- C— 180°, 
F = area of triangle ; 

then, by Solid Geometry, 

*=£?'*■ [50] 

II. When the three sides a, b, c, are given. 

A formula for computing the area is deduced as follows : 
From the first of [48], 

cosK^+-£) _ cosl(a + b) . 
cos (90° -£ (7) cosi<? 

whence, by the Theory of Proportions, 

coa$(A + B) — coa(90 o —$ C)__ co8%(a + b) — cosic / a ) 
cos|(^ + ^) + cos(90°-iC) coai(a + b) + coa±c 

Now, in § 35, the division of [23] by [22] gives 
cos A — cos B 



cos A-\- cos B 



- tan l(A + B) tan l(A-B), (b) 



in which for A and B we may substitute any other two angu- 
lar magnitudes, as for example, I ( A -f B) and (90 — £ C), or 
I (a -f- J) and J c. 

If we use in place of A and J5 the values \{A-\- B) and 
(90° — •§- C), the first side of equation (b) becomes 

cos±(A + B)- cos(90°- JO) . 
cos±(A + £) + cos(90°- IC) ' 



134 TRIGONOMETRY. 



and the second side becomes 

-tanKM + -^+90 o -|-C)tanKU + W-90°+^); 
or, 

- tan I (A + B - 0+ 180°) tan J (A + B + O- 180°). 

If we remember that E = A + B -{- C— 180°, and observe 
that 

ta.ni(A + B-C+180°)=t&n\(360 o -2C+A + B+C-180°) 
=ttmi(3W o -20+E) 
=tnn[90°-l(2O-E)] 
= cot i (20- E), 

it will be evident that equation (b) may be written 

cos i(^)-cos(90°-^) = __ ^^ 

cos J (^4+5) + cos (90°- JC) V ; ^ ; 

If we substitute, in equation (b), for A and B, the values 
£ (a + J) and \ c, and also substitute s for £ (a -f b -J- c) and 
s — <? for | (a -\- b — c), equation (b) will become 

cos \ (a-\-b) — cos \c . i - i / \ / -i \ 

\) \ [ — = - tan J stan *(*-*)• (d) 

cos 2 {a + o) -f- cos k c 

Comparing (a), (c), and (d), we obtain 

cot \ (20- E ) tan \ E = tan \ s tan \ (s - c). (e) 

By beginning with the second equation of [48], and treating 
it in the same way, we obtain as the result, 

tan i (20- E) tan \ E = tan£($ — a) tan £ (s - b). (f) 

By taking the product of (e) and (f), we obtain the elegant 
formula, 

tan 2 \ E = tan \ s tan § (s — a) tan \ (s — b) tan £(s — c), [51] 

which is known as l'Huilier's Formula. 

By means of it E may be computed from the three sides, 
and then the area of the triangle may be found by [50]. 






THE OBLIQUE SPHERICAL TRIANGLE. 



135 



III. In all other cases, the area may be found by first solv- 
ing the triangle so far as to obtain the angles or the sides, 
whichever may be more convenient, and then applying [50] 
or [51]. 



Example I. 



4 = 102° 14' 12" 
B = 54° 32' 24" 
0= 89° 5' 46" 



245° 52' 22" 
E= 65° 52' 22" 



180 c 



237142" 
648000" 



log 



log^ = 


- 5.37501 


7T 


= 4.68557 - 10 


648000 


0.06058 


F= 


= 1.1497iZ 2 



Example II. a = 133° 26' 19" 
6= 64° 50' 53" 
c = 144° 13' 45" 

2s = 342° 30' 57" 

s = 171° 15' 28.5" 

s-a = 37° 49' 9.5" 

s- b = 106° 24' 35.5" 

s-c= 27° 1'43.5" 



Hs-c) 



85°37 , 44" 
18° 54' 35" 
53° 12' 18" 
13°30 / 52" 





log tan 2 s = 


= 1.11669 


log 


tan i(s - 


-a) = 


= 9.53474 


log 


tan \ (s - 


-b) = 


= 0.12612 


log 


tan \ (s - 


-«)■ 


= 9.38083 



log tan 2 1 E-- 

logtan \E- 

\E- 

E- 



0.15838 
0.07919 
50° 11' 43" 
200° 46 '52" 



Exercise XXXI. 



1. Given A = 84° 20' 19", B = 27° 22' 40", (7= 75° 33'; 
find £= 26159", F= 0.12685 J? 2 . 

2. Given a = 69° 15' 6", b = 120° 42' 47", c = 159° 18' 33" ; 
find E= 216° 40' 18". 

3. Given a == 33° 1' 45", 6 = 155° 5' 18", (7=110° 10'; 
find ^=133° 48' 53". 

4. Find the spherical excess of a triangle on the earth's 
surface (regarded as spherical), if each side of the triangle is 
equal to 1°. 



See Wentworth & Hill's Tables, page 20, 



CHAPTEE VII. 



APPLICATIONS OF SPHERICAL TRIGONOMETRY. 



§ 61. Problem. 
To reduce an angle measured in space to the horizon. 
Let (Fig. 44) be the position of the observer on the 

ground, AOB = h the angle 
measured in space (for example, 
the angle between the tops of 
two church spires), OA 1 and OB' 
the projections of the sides of the 
angle upon the horizontal plane 
BR, AOA' = m and BOB' = n 
the angles of inclination of OA 
and OB respectively to the 
horizon. Required the angle 
A'OB' = x made by the projec- 
tions on the horizon. 
The planes of the angles of inclination AOA 1 and BOB' 
produced intersect in the line OC, which is perpendicular to 
the horizontal plane (Greom. § 475). 

From as a centre describe a sphere, and let its surface 
cut the edges of the trihedral angle = ABC in the points 
M, JVj and P. In the spherical triangle MNP the three 
sides MJSF=h, MP = 90 — m, NP = 90 — n, are known, and 
the spherical angle P is equal to the required angle x. 
From § 52 we obtain 




R 



Fig. 44. 



cos -J- x = Vcos s cos (s — h) sec m sec n, 



where J (m + n -f- h) = s. 



APPLICATIONS. 



137 



§ 62. Problem. 

To find the distance between two places on the earth's surface 
{regarded as spherical), given the latitudes of the places and 
the difference of their longitudes. 

Let M and N (Fig. 45) be the places ; then their distance 
MN is an arc of the great cir- 
cle passing through the places. 
Let P be the pole, AB the 
equator. The arcs MR and 
NS are the latitudes of the 
places, and the arc RjS, or the 
angle MPN, is the difference 
of their longitudes. Let MR 
= a, JSTS=b, RS=l; then in 
the spherical triangle MNP 
two sides, MP =90 -a, NP 
= 90 — b, and the included 
angle MPN=l, are given, and we have (from 

tan m = cot a cos Z, 

cos MN= sin a sec m sin (b -f- m). 

From these equations first find w, then the arc MJSF, and 
then reduce MNto geographical miles, of which there are 60 
in each degree. 




Fig. 45. 



54) 



§ 63. The Celestial Sphere. 

The Celestial Sphere is an imaginary sphere of indefinite 
radius, upon the concave surface of which all the heavenly 
bodies appear to be situated. 

The Celestial Equator, or Equinoctial, is the great circle in 
which the plane of the earth's equator produced intersects 
the surface of the celestial sphere. 

The Poles of the equinoctial are the points where the earth's 
axis produced cuts the surface of the celestial sphere. 



138 TRIGONOMETRY. 



The Celestial Meridian of an observer is the great circle in 
which the plane of his terrestrial meridian produced meets 
the surface of the celestial sphere. 

Hour Circles, or Circles of Declination, are great circles passing 
through the poles, and perpendicular to the equinoctial. 

The Horizon of an observer is the great circle in which a 
plane tangent to the earth's surface, at the place where he is, 
meets the surface of the celestial sphere. 

The Zenith of an observer is that pole of his horizon which 
is exactly above his head. 

Vertical Circles are great circles passing through the zenith 
of an observer, and perpendicular to his horizon. 

The vertical circle passing through the east and west points 
of the horizon is called the Prime Vertical; that passing 
through the north and south points coincides with the celestial 
meridian. 

The Ecliptic is a great circle of the celestial sphere, appar- 
ently traversed by the sun in one year from west to east, in 
consequence of the motion of the earth around the sun. 

The Equinoxes are the points where the ecliptic cuts the 
equinoctial. They are distinguished as the Vernal equinox 
and the Autumnal equinox; the sun in his annual journey 
passes through the former on March 21, and through the 
latter on September 21. 

Circles of Latitude are great circles passing through the 
poles of the ecliptic, and perpendicular to the plane of the 
ecliptic. 

The angle which the ecliptic makes with the equinoctial is 
called the obliquity of the ecliptic; it is equal to 23° 27', 
nearly, and is often denoted by the letter e. 

These definitions are illustrated in Figs. 46 and 47. In 
Fig. 46, AVBU is the equinoctial, P and P' its poles, NPZS 
the celestial meridian of an observer, NESWhis horizon, Z 
his zenith, M a star, PMP the hour circle passing through 
the star, ZMDZ 1 the vertical through the star. 



APPLICATIONS. 



139 



In Fig. 47, AVBTJ represents the equinoctial, EVFTJ the 
ecliptic, P and Q their respective poles, "Fthe vernal equinox, 
U the autumnal equinox, M a star, PMP the hour circle 
through the star, QMT the circle of latitude through the star, 
and ZTVP = e. 




8 A 




Fig. 47. 

The earth's diurnal motion causes all the heavenly bodies 
to appear to rotate from east to west at the uniform rate of 
15° per hour. If in Fig. 46 we conceive the observer 
placed at the centre 0, and his zenith, horizon, and celestial 
meridian fixed in position, and all the heavenly bodies rotat- 
ing around PP 1 as an axis from east to west at the rate of 15° 
per hour, we form a correct idea of the apparent diurnal 
motions of these bodies. When the sun or a star in its diur- 
nal motion crosses the meridian, it is said to make a transit 
across the meridian; when it passes across the part NWS 
of the horizon, it is said to set; and when it passes across the 
part NUjS, it is said to rise (the effect of refraction being 
here neglected). Each star, as M, describes daily a small 
circle of the sphere parallel to the equinoctial, and called the 
Diurnal Circle of the star. The diurnal circle is the smaller 
the nearer the star is to the pole ; and if there were stars at 
the poles P and P\ they would have no diurnal motion, To 



140 TRIGONOMETRY. 



an observer north of the equator, the north pole P is elevated 
above the horizon (as shown in Fig. 46) ; to an observer south 
of the equator, the south pole I* is the elevated pole. 

§ 64. Spherical Co-ordinates. 

Several systems of fixing the position of a star on the sur- 
face of the celestial sphere at any instant are in use. In each 
system a great circle and its pole are taken as standards of 
reference, and the position of the star is determined by means 
of two quantities called its spherical co-ordinates. 

I. If the horizon and the zenith are chosen, the co-ordinates 
of the star are called its altitude and its azimuth. 

The Altitude of a star is its angular distance, measured on 
a vertical circle, above the horizon. The complement of the 
altitude is called the Zenith Distance. 

The Azimuth of a star is the angle at the zenith formed by 
the meridian of the observer and the vertical circle passing 
through the star, and is measured therefore by an arc of the 
horizon. It is usually reckoned from the north point of the 
horizon in north latitudes, and from the south point in south 
latitudes ; and east or west according as the star is east or 
west of the meridian. 

II. If the equinoctial and its pole are chosen, then the posi- 
tion of the star may be fixed by means of its declination and 
its hour angle. 

The Declination of a star is its angular distance from the 
equinoctial, measured on an hour circle. The angular dis- 
tance of the star, measured on the hour circle, from the elevated 
pole is called its Polar Distance. 

The declination of a star, like the latitude of a place on the 
earth's surface, may be either north or south ; but, in practical 
problems, while latitude is always to be considered positive, 
declination, if of a different name from the latitude, must be 
regarded as negative. 



APPLICATIONS. 141 



If the declination is negative, the polar distance is equal 
numerically to 90° + the declination. 

The Hour Angle of a star is the angle at the pole formed by 
the meridian of the observer and the hour circle passing 
through the star. On account of the diurnal rotation, it is 
constantly changing at the rate of 15° per hour. Hour angles 
are reckoned from the celestial meridian, positive towards the 
west, and negative towards the east. 

III. The equinoctial and its pole being still retained, we 
may employ as the co-ordinates of the star its declination and 
its right ascension. 

The Eight Ascension of a star is the arc of the equinoctial 
included between the vernal equinox and the point where the 
hour circle of the star cuts the equinoctial Eight ascension is 
reckoned from the vernal equinox eastward from 0° to 360°. 

IV. The ecliptic and its pole may be taken as the standards 
of reference. The co-ordinates of the star are then called its 
latitude and its longitude. 

The Latitude of a star is its angular distance from the eclip- 
tic measured on a circle of latitude. 

The Longitude of a star is the arc of the ecliptic included 
between the vernal equinox and the point where the circle of 
latitude through the star cuts the ecliptic. 

In problems it is useful to employ certain letters to denote 
these various co-ordinates. For a star M (Fig. 46), let 
I = latitude of the observer, 
h = DM = the altitude of the star, 
z = ZM = the zenith distance of the star, 
a = ZPZM= the azimuth of the star, 
t = ZZPM= the hour angle of the star, 
d == R M = the declination of the star, 
p = PM = the polar distance of the star, 
r = VP = the right ascension of the star, 
u= MT (Fig. 47) = the latitude of the star, 
v= FT (Fig. 47) = the longitude of the star. 



142 



TRIGONOMETRY. 



In many problems, a simple way of representing the mag- 
nitudes involved, is to project the sphere on the plane of the 

horizon, as shown in Fig. 48. 




NESW is the horizon, Z 
the zenith, NZ8 the meridian, 
WZUthe prime vertical, WAE 
the equinoctial projected on the 
plane of the horizon, P the ele- 
vated pole, M a star, DM its 
altitude, ZM its zenith dis- 
tance, APZM its azimuth, 
MR its declination, PM its 
polar distance, /.ZPM its hour 
angle. 



§ 65. The Astronomical Triangle. 

The triangle ZPM (Figs. 46 and 48) is often called the 
astronomical triangle, on account of its importance in prob- 
lems in Nautical Astronomy. 

The side PZ is equal to the complement of the latitude of 
the observer. For (Fig. 46) the angle ZOB between the 
zenith of the observer and the celestial equator is obviously 
equal to his latitude, and the angle POZ is the complement 
of ZOB. The arc NP being the complement of PZ, it follows 
that the altitude of the elevated pole is equal to the latitude of 
the place of observation. 

The triangle ZPM then (however much it may vary in 
shape for different positions of the star M ), always contains 
the following five magnitudes : 

PZ= co-latitude of observer = 90°— I, 

ZM= zenith distance of star = z, 
PZM= azimuth of star = a, 

PM= polar distance of star =p, 
ZPM= hour angle of star = t. 



APPLICATIONS. 143 



A very simple relation exists between the hour angle of the 
sun and the local (apparent) time of day. Since the hourly 
rate at which the sun appears to move from east to west is 
15°, and it is apparent noon when the sun is on the meridian 
of a place, it is evident that if hour angle = 0°, 15°, — 15°, etc., 
time of day is noon, 1 o'clock p.m., 11 o'clock A.M., etc. 

In general, if t denote the absolute value of the hour angle, 

time of day = — p.m., or 12 a.m., 

J 15 15 

according as the sun is west or east of the meridian. 

§ 66. Problem. 

Given the latitude of the observer and the altitude and azimuth 
of a star, to find its declination and its hour angle. 

In the triangle ZPM (Fig. 48), 

given PZ= 90° — 1= co-latitude, 

ZM= 90° - h = co-altitude, 

ZPZM=a = azimuth; 

to find PM= 90° - d = polar distance, 

ZZPM= t = hour angle. 

Draw MQJL NS, and put ZQ = m, 

then, if a < 90°, PQ = 90° - (I + m), 

and if a > 90°, PQ = 90° -\l - m) ; 

and, by Napier's Rules, 

cos a = ± tan m tan A, 
sin d = cos PQ cos MQ, 
sin A = cosm cos MQ; 

whence, tanm = ± cot A cos a, 

sin d = sin h sin (I ± rri) sec ra, 

in which the — sign is to be used if a > 90°. The hour, angle 
may then be found by means of [43], whence we have 

sin t = sin a cos h sec d. 



144 



TRIGONOMETRY. 




§ 67. Problem. 

To find the hour angle of a heavenly body when its declina- 
tion, its altitude, and the lati- 
tude of the place are known. 

In the triangle ZPM (Fig. 

PZ= 90° -I, 
PM=90°-d = p, 
ZM=90°-h; 

required 

Z ZPM= t. 
If, in the first formula of [46], 

sin \ A = Vsin (s — b) sin (s — c) esc b esc c, 
we put 

A = t, a = 90°-h, b=p, c = 90°-l, 
we have 

8 -b = 90°-i(l+p + h) 1 s-c==i(l+p-h), 

and the formula becomes 

sin it = ± [cos J (l+p + h) sin £ (l-\-p — h) sec I cscp]* 

in which the — sign is to be taken when the body is east of 
the meridian. 

If the body is the sun, how can the local time be found 
when the hour angle has been computed ? (See § 65.) 



APPLICATIONS. 145 



§ 68. Problem. 

To find the altitude and azimuth of a celestial body, when its 
declination, its hour angle, and the latitude of the place are 
known. 

In the triangle ZPM (Fig. 49), 

given PZ=90°-l, 

PM= 90° ~i—p t 

Z.ZPM=t; 
required ZM= 90° -h, 

ZPZM= a. 

Here there are given two sides and the included angle. 
Placing PQ = m, and proceeding as in § 66, we obtain 

tan m — cot d cos t, 

sin h = sin {I -J- m) sin d sec ra, 

tan a = sec (I -f- m) tan t sin m, 

in the last of which formulas a must be marked E. or W., to 
agree with the hour angle. 

§ 69. Problem. 

To find the latitude of the place when the altitude of a celes- 
tial body, its declination, and its hour angle are known. 

In the triangle ZPM (Fig. 49), 
given ZM= 90° -h, 

PM= 90° -d, 
ZZPM=t; 
required PZ= 90° - I. 

Let PQ = m, ZQ = n. 



146 



TRIGONOMETRY. 



Then, by Napier's Rules, 
N 



cos t = tan m tan d, 
sin h = cosn cos MQ, 
sin d = cosra cos MQ ; 

whence, 

tan m = cot d cos £, 

cos n = cos ra sin h esc c?, 

and it is evident from the fig- 
ure that 

Z=90°-(?tt±n), 

in which the sign -f- or the sign 

— is to be taken according as 

the body and the elevated pole 

are on the same side of the prime vertical or on opposite 

sides. 

In fact, both values of I may be possible for the same alti- 
tude and hour angle; but, unless n is very small, the two 
values will differ largely from each other, so that the observer 
has no difficulty in deciding which of them should be taken. 




§ 70. Problem. 

Given the declination, the right ascension of a star, and the 

obliquity of the ecliptic, to find 
the latitude and the longitude of 
the star. 

Let M (Fig. 50) be the star, 
Pbe the pole of the equinoctial, 
and Q the pole of the ecliptic. 

Then, on the triangle PMQ, 

given PQ = e = 2Z°2T, 
P3f=<d0°-d, 
ZMPQ= 90° -|-r(sefe Fig. 47); 

and ZP<UT= 90° -?;(see Fig. 47). 




Fig. 50. 



required QM=90 C 



APPLICATIONS. 147 



In this case, also, two sides and the included angle are given. 
Draw MSA.PQ, and meeting it produced at S, and let 
PII=n. 

By Napier's Rules, 

sin r = tan n tan d, 
sin u = cos {e -f- n) cos MS, 
sin d = cos n cos MS, 
sin (<? -f- w) = tan v tan Jffl", 
sin n = tanr tan MS; 
whence, tan n = cot d sin r, 

sin u = sin c? cos (e -f- 92) sec w, 
tan v = tan r sin (e -f- w) esc n. 

To avoid obtaining m from its sine we may proceed as fol- 
lows : 

From the last two equations we have, by division, 

sin u = tan v cot (e -f- n) sin d cot r tan n. 

By taking MS as middle part, successively, in the triangles 
MQSsmd MPS, we obtain 

cos w cos v — cos c? cos r ; 

whence, cos w = sec v cos c? cosr. 

From these values of sin u and cosw we obtain, by division, 

tan t/. = sin v cot(e -j- n) tan d esc r tanw. 

From the relation 

sin r = tan n tan c?, 

it follows that tan d esc r tan w-1. 

Therefore tan w = sin v cot (e -f- n), 

a formula by which u can be easily found after v has been 
computed. 



148 TRIGONOMETRY. 



Exercise XXXII. 

1. Find the dihedral angle made by the lateral faces of a 
regular ten-sided pyramid; given the angle A =18°, made at 
the vertex by two adjacent lateral edges. 

2. Through the foot of a rod which makes the angle A with 
a plane, a straight line is drawn in the plane. This line makes 
the angle B with the projection of the rod upon the plane. 
What angle does this line make with the rod ? 

3. Find the volume V of an oblique parallelopipedon ; 
given the three unequal edges a, b, c, and the three angles 
I, m, n, which the edges make with one another. 

4. The continent of Asia has nearly the shape of an equi- 
lateral triangle, the vertices being the East Cape, Cape Romania, 
and the Promontory of Baba. Assuming each side of this tri- 
angle to be 4800 geographical miles, and the earth's radius to 
be 3440 geographical miles, find the area of the triangle : (i.) 
regarded as a plane triangle; (ii.) regarded as a spherical 
triangle. 

5. A ship sails from a harbor in latitude I, and keeps on 
the arc of a great circle. Her course (or angle between the 
direction in which she sails and the meridian) at starting is a. 
Find where she will cross the equator, her course at the equa- 
tor, and the distance she has sailed. 

6. Two places have the same latitude I, and their distance 
apart, measured on an arc of a great circle, is d. How much 
greater is the arc of the parallel of latitude between the places 
than the arc of the great circle? Compute the results for 
Z = 45°, d=90°. 

7. The shortest distance d between two places and their 
latitudes I and V are known. Find the difference between 
their longitudes. 

8. Given the latitudes and longitudes of three places on the 
earth's surface, and also the radius of the earth ; show how to 
find the area of the spherical triangle formed by arcs of great 
circles passing through the places. 



APPLICATIONS. 149 



9. The distance between Paris and Berlin (that is, the arc 
of a great circle between these places) is equal to 472 geo- 
graphical miles. The latitude of Paris is 48° 30' 13"; that of 
Berlin, 52° 30' 16". When it is noon at Paris what time is it 
at Berlin ? 

Note. Owing to the apparent motion of the sun, the local time over 
the earth's surface at any instant varies at the rate of one hour for 15° 
of longitude ; and the more easterly the place, the later the local time. 

10. The altitude of the pole being 45°, I see a star on the 
horizon and observe its azimuth to be 45° ; find its polar 
distance. 

11. Given the latitude I of the observer, and the declina- 
tion d of the sun ; find the local time (apparent solar time) of 
sunrise and sunset, and also the azimuth of the sun at these 
times (refraction being neglected). When and where does the 
sun rise on the longest day of the year (at which time d = 
+23° 27') in Boston (J = 42° 21'), and what is the length of 
the day from sunrise to sunset? Also, find when and where 
the sun rises in Boston on the shortest day of the year (when 
d=> — 23° 27'), and the length of this day. 

12. When is the solution of the problem in Example 11 im- 
possible, and for what places is the solution impossible ? 

13. Given the latitude of a place and the sun's declination ; 
find his altitude and azimuth at 6 o'clock a.m. (neglecting re- 
fraction). Compute the results for the longest day of the year 
at Munich (1= 48° 9'). 

14. How does the altitude of the sun at 6 a.m. on a given 
day change as we go from the equator to the pole ? At what 
time of the year is it a maximum at a given place ? (Given 
sin h = sin I sin d.) 

15. Given the latitude of a place north of the equator, and 
the declination of the sun ; find the time of day when the sun 
bears due east and due west. Compute the results for the 
longest day at St. Petersburg (1= 59° 56'). 



150 TKIGONOMETRY. 



16. Apply the general result in Example 15 (cosi( = cotZ 
tanc?) to the case when the days and nights are equal in 
length (that is, when d = 0°). Why can the sun in summer 
never be due east before 6 A.M., or due west after 6 p.m. ? 
How does the time of bearing due east and due west change 
with the declination of the sun ? Apply the general result to 
the cases where l< d and 1= d. What does it become at the 
north pole ? 

17. Given the sun's declination and his altitude when he 
bears due east ; find the latitude of the observer. 

18. At a point in a horizontal plane MJN"a, staff OA is 
fixed, so that its angle of inclination AOB with the plane 
is equal to the latitude of the place, 51° 30' N., and the direc- 
tion OB is due north. What angle will OB make with the 
shadow of OA on the plane, at 1 p.m., when the sun is on the 
equinoctial ? 

19. What is the direction of a wall in latitude 52° 30' N. 
which casts no shadow at 6 a.m. on the longest day of the 
year? 

20. At a certain place the sun is observed to rise exactly in 
the north-east point on the longest day of the year ; find the 
latitude of the place. 

21. Find the latitude of the place at which the sun sets at 
10 o'clock on the longest day. 

22. To what does the general formula for the hour angle, 
in § 67, reduce when (i.) h = 0°, (ii.) 1=0° and d=0°, (iii.) 
lord = 90°? 

23. What does the general formula for the azimuth of a 
celestial body, in § 68, become when^ = 90° = 6 hours ? 

24. Show that the formulas of § 69, if ^ = 90°, lead to the 
equation sin Z— sin h cscc?; and that if d=0°, they lead to 
the equation cos I = sin h sec t. 

25. Given latitude of place 52° 30' 16", declination of star 
38°, its hour angle 28° 17' 15" ; find its altitude. 



APPLICATIONS. 151 



26. Given latitude of place 51° 19' 20", polar distance of 
star 67° 59' 5', its hour angle 15° 8' 12" ; find its altitude and 
its azimuth. 

27. Given the declination of a star 7° 54', its altitude 22° 
45' 12", its azimuth 129° 45' 37" ; find its hour angle and the 
latitude of the observer. 

28. Given the longitude v of the sun, and the obliquity of 
the ecliptic e = 23°27'; find the declination d, and the right 
ascension r. 

29. Given the obliquity of the ecliptic e= 23° 27', the lati- 
tude of a star 51°, its longitude 315° ; find its declination and 
its right ascension. 

30. Given the latitude of place 44° 50' 14", the azimuth of 
a star 138° 58' 43", and its hour angle 20° ; find its declination. 

31. Given latitude of place 51° 31' 48", altitude of sun west 
of the meridian 35° 14' 27", its declination +21° 27' ; find the 
local apparent time. 

32. Given latitude of place /, the polar distance p of a star, 
and its altitude h ; find its azimuth a. 







APPENDIX. 



FORMULAS. 



1. sin 2 A + cos 2 J. = 1. 1 ' 
sin A 



2. tan A 



cos A 



{sin A X esc A = 1.1 
cos A x sec ^4. = 1. 
tan, J. X cot^i = 1. 



I §5. 



4. sin (x + y) = sin a? cos y -j- cos a; sin y. 

5. cos (x + y) = cos a; cos y — sin a; sin y. 

6. tan(^ + y)= 1 tana: + tany - 

1 — tan x tan y 

rr i. / i \ cot a? cot v — 1 

7. cot (a; + y) = — - — -*— — 

cot a;-}- coty 

8. sin (x—y)= sin x cos y — cos a? sin y. 

9. cos (a; — y) ,= cos a; cos y -f sin a; sin y . 



I- § 3i. 



10. tan (x — y) 



tan x — tan y 
1 -j- tan x tan y 



§32. 



n , / x cot a? coty + 1 

11. cot (x — y) = 2— ! — 

cot y — cot x 



12. sin 2 a; = 2 sin a? cos x. 



13. cos 2 a; = cos 2 a? — sin 2 £. 



§33. 



154 

14. tan 2a; 

15. cot 2x 

16. sin | 



FORMULAS. 



2 tan a; 
1 — tan 2 a: 

cot 2 ff -- 1 
2 cot a: 



§33. 



-~f 



cosz 



17. cos I 



U^±^- 



-f cosz 



18. tan |z= ±\k — - 

\ 1 -f- cos z 



19. cot | 2 



=^l 



+ COSZ 



COS 2 



§34. 



21. 
22. 
23. 

24. 
25. 
26. 
27. 



sin J. + sin B = 2 sin i (A + .£) cos \{A-B). 
sin J. - sin B = 2 cos \ (A + £) sin I (A - B). 
cos.4 + cos .5 -= 2 cos £ (A + B) cos J (J. - B). 
cgsA--cosB = -2 sin J (A + 5) sin | (A - B). 

sin J. + sin B _ tan $Q* + .5) 
sin ^4 — sin i? tan 5 ( J. — i?) 

a sin y! c D /> 
5- = - — -• §36. 

sinii 

a* = 6* + c*-2ftccos^. §37. 

a — b _ tan £ (^4 — B) « 00 
a + 6~tan£(^ + ^) ' 



sin £ ^4 



a' 



(s-b)(s-c) 



be 



§43. 



FORMULAS. 



155 



29. cos £ A 



s(s — a) 



\ j c 



30. ten^=J('- 6 > ( '- c >- 

\ s (s — a) 



31. 



#- 



)(s _ & )( S _ c) _ 



r. 



§43. 



32. tan i A = 

s — a 

33. F= I ac sin B. 

34. JF= Vs (*-«)(* -6) («-<?)• 



35. F- 



abc 



36. JF= %r(a + b + c) = rs. 



§44. 



Spherical Trigonometry. 
37. cos c = cos a cos 6. 
sin a = sine sin A. 



38. 

39. | 

40. | 



sin b = sin c sin i?. 

cos A = tan 6 cot c. 
cos j5 = tan a cot e. 

cos ^4 = cos a sin i?. 
cos B = cos 6 sin A. 



41. ( sin 6 

I sin a 



sin 6 = tana cot A. 
tan £ cot B. 



■ §46. 



42. cos c = cot ^L cot B. 

sin a sin B =■ sin & sin A. 

43. <! sin a sin = sin c sin A. 
sin 6 sin C — sin c sin B. 



5J. 



156 



FORMULAS. 



cos a = cos b cos c + sin b sin c cos A. 

44. -{ cos J = cos a cos c -f- sin a sin c cos B. 
cos c = cos a cos 5 -j- sin a sin 6 cos C. 

cos .4 = — cos B cos (7 -f- sin B sin (7 cos a. 

45. { cos .5 = — cos .4 cos (7 -f sin .4 sin C cos &. 
cos O = — cos .4 cos B -j- sin J. sin -5 cos e. 



§51. 



sin \ A = Vsin (s — "&) sin (s — c) esc 5 esc c. 

cos J .4 = Vsin s sin (s — a) esc 5 esc <?. 

tan ? -4 = Vcsc s esc (s — a) sin (s — 5) sin (s — c). 

sin £ a = V— cos # cos (# — A) esc i? esc (7. 
±7 ■ cos J a =Vcos (#- 5) cos(/S'- <7) esc B esc £ 
tan \ a =V— cos/S'cos (#— .4) sec (S—B) sec (S—C). 



§52. 



- cos £ (.4 + -5) cos J c = cos I (a-\-b) sin £ (7. 
sin £ (.4 + B) cos J c = cos J (a — b) cos £ (7. 
cos i (A — i?) sin i c = sin £ (a + &) sin J (7. 
sin j ( J. — B) sin J c = sin i (a — b) cos J C. 

cos £ (a -f 6) 



49 J 



tan * (.4 - 5) 



_ sin ^ (a — 5) 



cot J (7. 



sin I (a + 5) 

V 7 cos£(^ + ^) 

sin J (J. -f i?) 



50. i^ 



J£ 



180 c 



i? 2 . 



§53. 



51. tan 2 \ E= tan J (s — a) tan £ (s — b) tan £ (s — c). 



§60. 



FORMULAS. 157 



Prof. Blakslee's construction by which the direction ratios 
for plane right triangles give directly from a figure the analo- 
gies for a right trihedral or for a right spherical triangle. 




The construction consists of two parts. 

(a) Lay off from the vertex V& unit's distance on each edge. 

(b) Pass through the three extremities of these distances 
three planes perpendicular to one of the edges, as VA. Now 
these three parallel planes will cut out three similar right 
triangles. The first being constructed in either of the two 
usual ways, the construction of the others is evident. 

Since the plane angles A ly A 2 , A 3 all equal the dihedral A, 
and the nine right triangles in the three faces give the values 
in the figure, we have : 

(1) sin A = sin a : sin h ; similarly, sin B — sin b : sin h. 

(2) cos A = tan b : tan h ; similarly, cos B = tan a : tan h. 

(3) tan A = tana : sin b ; similarly, tan B = tan b : sin a. 

(4) cos h = cos a cos b ; (by 3) = cot A cot B. 

(5) sin A = cos B : cos b ; sin B = cos A : cos a. • 



Note. If a sphere of unit radius be described about V as a centre, the 
three faces will cut out a right spherical triangle, having the sides a, b, 
and h, and angles A, B, and H. The above formulas are thus seen to 
be the analogies of: 



158 



FORMULAS. 



(1) sin A = a : h ; sin B = b : h. 

(2) cos A = b : h ; cos B = a: h. 

(3) tan^. = a : 6 ; tan 5 = 5 : a. 

(4) A 2 = a 2 + b 2 ; 1 = sin 2 + cos 2 ; 1 

(5) sin A = cos B ; sin 5 = cos .A. 

Napier's rules give only the following, which follow from the analo- 
gies as numbered : 

By | sin a = sin A sin h = tan b cot B 
(1) I sin b = sin i? sin h = tan a cot A 
/-x f cos JL = sin B cos a = tan b cot A 
1 cos B = sin A cos 6 = tan a cot A 
(4) { cos h — cos a cos b = cot J. cot B } (4) 



cot A cot 5. 



(3) 
(2) 



The Gauss Equations. 




cos £( A -f -#) cos ^ c = cos l (a -f- 5) sin J C 
sin \ (A + -5) cos \ c = cos |- (a — 6) cos £ C 
cos l( J. — B) sin i c = sin %(a + b) sin -§- (7. 
sin \{A — B) sin |c— sin J (a — 6) cos £ (7. 

Rule I. sin in (I.) gives — in (3), and conversely, 
cos in (I.) gives -f- in (3), and conversely. 

Rule II. Functions have same names in (2) and (3). 
Functions have co-names in (4) and (1). 



SUKVEYING AND NAVIGATION. 



BY 

G. A. WENTWOBTH, A.M., 

PEOFESSOR OF MATHEMATICS IN PHILLIPS EXETER ACADEMY. 



>**c 



BOSTON: 

PUBLISHED BY GINN & COMPANY. 

1887. 



Entered, according to Act of Congress, in the year 1883, by 

G. A. WENTWORTH, 
in the Office of the Librarian of Congress, at "Washington. 



J. S. Cushing & Co., Pbinteks, Boston, Mass. 



PREFACE. 



THE object of this brief work on Surveying and Navigation is 
to present these subjects in a clear and intelligible way, accord- 
ing to the best methods in actual use; and also to present them in 
so small a compass that students in general may find the time to 
acquire a competent knowledge of these very interesting and impor- 
tant studies. 

The author is under special obligation to Professor James L. Pat- 
terson of Lawrenceville, N.J., who has done a large share of the 
work on both the Surveying and the Navigation. 

G. A. WENTWORTH. 
Phillips Exetee Academy, 
September, 1883. 



CONTENTS. 



SURVEYING. 

CHAPTER I. Definitions. Instruments and Their Uses: 

Definitions, 1 ; instruments for measuring lines, 2 ; chaining, 2 ; 
obstacles to chaining, 4 ; the measurement of angles, 6 ; the vernier 
compass, 7; uses of the compass, 9; verniers, 11; the surveyor's 
transit, 15 ; uses of the transit, 16 ; the theodolite, 16 ; the railroad 
compass, 16 ; plotting, 19. 

CHAPTER II. Land Surveying: 

Determination of areas, 21 ; rectangular surveying, 25 ; field notes, 
computation, and plotting, 26 ; supplying omissions, 30 ; irregular 
boundaries, 30; obstructions, 30; modification of the rectangular 
method, 33 ; variation of the needle, 34 ; methods of establishing a 
true meridian, 36 ; dividing land, 39 ; United States public lands, 42; 
Burt's solar compass, 43; laying out the public lands, 45; Plane- 
table surveying, 47 ; the three-point problem, 52. 

CHAPTER III. Triangulation : 

Introductory remarks, 53 ; the measurement of base lines, 54 ; 
the measurement of angles, 55. 

CHAPTER IV. Levelling: 

Definitions, 56 ; the Y level, 57; the levelling-rod, 57; difference 
of level, 58; levelling for section, 61 ; substitutes for the Y level, 64; 
topographical levelling, 66. 

CHAPTER V. Railroad Surveying: 

General remarks, 68 ; cross-section work, 68 • railroad curves, 69. 



VI 



CONTENTS 



NAVIGATION. 

CHAPTER I. Definitions. Couese and Distance: 

Definitions, 1 ; the mariner's compass, 2 ; deviation, 3 ; the 
wind, 6; leeway, 7; reducing the courses, 7; the log, 9; currents, 13. 

CHAPTER II. The Sailings: 

Definitions, 14; notation, 14; plane sailing, 15; parallel sail- 
ing, 19 ; middle latitude sailing, 22 ; Mercator's sailing, 27 ; traverse 
sailing, 33 ; the day's work, 36 ; great circle sailing, 50. 



CHAPTER III. Nautical Asteonomy: 

The observed altitude of a heavenly body, 57 ; corrections of the 
observed altitude, 59 ; time, 66 ; longitude and time, 67 ; Greenwich 
date, 68 ; the chronometer, 69 ; the nautical almanac, 71 ; to find the 
hour angle of a heavenly body, 74 ; to find the local time, 74 ; lati- 
tude by meridian observations, 77 ; latitude by ex-meridian altitude 
of the sun, 81 ; longitude by chronometer, 83 ; deviation by time 
azimuths, 91. 



SURVEYING. 

CHAPTER I. 
DEFINITIONS. INSTRUMENTS AND THEIR USES. 



§ 1. Definitions. 

Surveying is the art of determining and representing dis- 
tances, areas, and the relative position of points upon the 
surface of the earth. 

In plane surveying, the portion surveyed is considered as a 
plane. 

In geodetic surveying, the curvature of the earth is re- 
garded. 

A Plumb-Line is a cord with a weight attached and freely 
suspended. 

A Vertical Line is a line having the direction of the plumb- 
line. 

A Vertical Plane is a plane embracing a vertical line. 

A Horizontal Plane is a plane perpendicular to a vertical line. 

A Horizontal Line is a line in a horizontal plane. 

A Horizontal Angle is an angle the sides of which are in a 
horizontal plane. 

A Vertical Angle is an angle the sides of which are in a 
vertical plane. If one side of a vertical angle is horizontal, 
and the other ascends, it is an angle of elevation ; if one side 
is horizontal, and the other descends, it is an angle of 
depression. 

The Magnetic Meridian is the direction which a bar-magnet 
assumes when freely supported in a horizontal position. 



SURVEYING. 



The Magnetic Bearing of a line is the angle it makes with 
the magnetic meridian. 

Surveying commonly comprises three distinct operations; 
viz. : 

1. The Field Measurements, or the process of determining by 
direct measurement certain lines and angles. 

2. The Computation of the required parts from the measured 
lines and angles. 

3. The Plotting, or representing on paper the measured and 
computed parts in relative extent and position. 



THE MEASUREMENT OF LINES. 
§ 2. Instruments for Measuring Lines. 

The Gunter's Chain is generally employed in measuring land. 
It is 4 rods, or 66 feet, in length, and is divided into 100 links. 
Hence, links may be written as hundredths of a chain. 

The Engineer's Chain is employed in surveying railroads, 
canals, etc. It is 100 feet long, and is divided into 100 links. 

A Tape Measure, divided into feet and inches, is employed 
in measuring town-lots, cross-section work in railroad survey- 
ing, etc. 

In the United States Coast and Geodetic Survey, the meter 
is the unit ; and, when great accuracy is required, rods placed 
end to end, and brought to a horizontal position by means of 
a spirit-level, are employed in measuring lines. 

§ 3. Chaining. 

Eleven tally-pins of iron or steel are used in chaining ; also, 
one or more slender poles shod with iron, and bearing a flag. 

A forward chainman, or leader, and a hind chainman, or 
follower, are required. A flag- staff having been placed at the 
farther end of the line, or at some point in the line visible 



CHAINING. 



from the beginning, the follower takes one end of the chain, 
and a pin which he thrusts into the ground at the beginning 
of the line. The leader moves forward in the direction of 
the flag-staff, with the other end of the chain and the re- 
maining ten pins, until the word "halt" from the follower 
warns him that he has advanced nearly the length of the 
chain. 

At this signal he stops, and the follower, meanwhile hav- 
ing placed his end of the chain at the beginning of the line, 
directs the leader by the words "right" or "left" until the 
chain is exactly in line with the flag-staff. This being accom- 
plished, and the chain stretched tightly in a horizontal posi- 
tion, the follower says, "down." The leader then puts in 
a tally-pin exactly at the end of the chain, and answers, 
"down"; after which the follower withdraws the pin at the 
beginning of the line, and the chainmen move forward until 
the follower nears the pin set by the leader. The follower 
again says, "halt," and the operation just described is re- 
peated. This process is continued until the end of the line 
is reached. 

If the tally-pins in the hands of the leader are exhausted 
before the end of the line is reached, when he has placed the 
last pin in the ground, he waits until the follower comes' up 
to him. The follower gives the leader the ten pins in his 
hand, and records the fact that ten chains have been meas- 
ured. The measuring then proceeds as before. If the distance 
from the last pin to the end of the line is less than a chain, 
the leader places his end of the chain at the end of the line, 
and the follower stretches tightly such a part of the chain 
as is necessary to reach to the last pin, and the number of 
links is counted. The number of whole chains is indicated by 
the number of pins in the hands of the follower, the last pin 
remaining in the ground. 

In measuring, the chain must be held in a horizontal posi- 
tion. If the ground slopes, one end of the chain must be 
raised until the horizontal position is attained. By means of 



SURVEYING. 



a plumb-line, or a slender staff, or, less accurately, by drop- 
ping a pin (heavy end downwards), the point vertically under 
the raised end of the chain may be determined. If the slope 
is considerable, half a chain or less may be used. 

To construct a perpendicular with a chain : 

1. When the point through which the perpendicular is to 
pass is in the line : 

Let AB (Fig. 1) represent the line, and P the point. Measure from 

P to the right or left, PC = 40 links, 

D and place a stake at C. Let one end 

s'\ of the chain be held at P, and the 

,' end of the eightieth link at C; then, 

/' taking the chain at the end of the 

<1 J thirtieth link from P. draw it so that 

AC P B 

Fj ( the portions DC and DP are tightly 

stretched, and place a stake at D. 

DP will be the perpendicular required. (Why ?) 

2. When the point is without the line : 

Let AB (Fig. 1) be the line, and D the point. Take C any point in 
the line, and stretch the chain between D and 0; then, let the middle of 
the part of the chain between C and D be held in place, and swing the 
end at D around until it meets the line in P. DP will be the perpen- 
dicular required. (Why ?) 

§ 4. Obstacles to Chaining. 

1. When a tree, building, or other obstacle is encountered 
in measuring or extending a line, it may be passed by an off- 
set in the following manner : 

Let O (Fig. 2) represent a building on the line AD. At B erect BE 

perpendicular to 

1 AB; at E erect 

i EF perpendicu- 

E ' . „ r lar to BE; at F 

Fig. 2. ■ 

erect FG = BE 
perpendicular to EF; then, CD perpendicular to FG will be in the 
required line, and AB + EF+ CD = AD. By constructing two other per- 
pendiculars, B"E' and F'C, the accuracy of the work will be increased. 



<& 



OBSTACLES TO CHAINING. 



2. To measure across a body of water : 

Let it be required to measure the line ABCD (Fig. 3) crossing a river 
between B and C. Measure BE = 400 links ; at E erect the perpendi- 
cular EF= 600 links ; at B erect the perpendicular BG = 300 links. 
Place a stake at C, the intersection of AD and FO beyond the river- 




Fig. 3. 

Then BQ= 400 links. For, by similar triangles, EF-. BG-..CE-. CB. 
But EF=2BG; hence, CE=2CB, and C5 = ££'=400 links. EG 
and FG should be measured, in order to test the accuracy of the work. 
EG = FG = 500 links. 

Instead of the above distances, any convenient distances may be taken. 
For, if EF= 2BG, then CB 



BE, and EG = FG 



-.VeIF+bg 2 . 



3. To measure a line the end of which is invisible from the 
beginning, and intermediate points unknown : 



Fig. 4. 

Let AB (Fig. 4) represent the line. Set up a flag-staff at D, beyond 
B and visible from A. From B let fall BC perpendicular to AD. Meas- 
ure AC and B O. Then __ 

ab=-Vac 2 + bc*. 

To find intermediate points on AB : 

At any point E on A C erect i£F perpendicular to AC, and determine 
EG by the proportion AC-.CB-.: AE: EG. G will be a point on AB 

The line AD is called a Kandom Line. 



SURVEYING. 



THE MEASTJKEMENT OF ANGLES. 
§ 5. The Vernier Compass.* 

The Vernier Compass is shown on the following page. 

The compass circle is divided into half-degrees, and is fig- 
ured from 0° to 90° each way from the north and south points. 
In the centre of the compass circle is the pivot which supports 
the magnetic needle. The needle may be lifted from the pivot 
by a spring and pressed against the glass covering of the 
compass circle, when the instrument is not in use. The main 
plate moves around the compass circle through a small arc, 
read by the vernier, for the purpose of allowing for the varia- 
tion of the needle (§ 23). The sight standards at the extremi- 
ties of the main plate have fine slits nearly their whole length, 
with circular openings at intervals; on the edges of the north 
standard tangent scales are placed ; and on the outside of the 
south standard two eye-pieces at the same distance from the 
main plate as the zeros of the tangent scales, respectively. 
The telescopic sight (a recent improvement by the Messrs. 
Gurley), consists of a small telescope attached to the south 
standard. The main plate is furnished with two spirit levels 
at right angles, and turns horizontally upon the upper end of 
the ball spindle, the lower end of which rests in a spherical 
socket in the top of the tripod or Jacob's staff which sup- 
ports the instrument. From the centre of the plate at the 
top of the tripod a plumb-bob is suspended by which the 
centre of the compass can be placed directly oyer a definite 
point on the ground. 

* The instruments described on this and the following pages are adjusted 
by the maker. If they should require readjustment, full directions will be 
found in the manual furnished with the instruments. 

The manual published by Messrs. W. & L. E. Gueley, Troy, N.Y., has 
been freely used, by permission, in describing these instruments. 



INSTRUMENTS AND THEIR USES. 



§ 6. Uses op the Compass. 

To take the bearing* of a line. Place the instrument so that 
the bob will be directly over one end of the line, and level by- 
pressing with the hands on the main plate until the bubbles 
are brought to the middle of the spirit levels. Turn the south 
end of the instrument toward you, and sight at the flag-staff 
at the other end of the line. Head the bearing from the north 
end of the needle. First, write N. or S. according as the 
north end of the needle is nearer N. or S. of the compass cir- 
cle ; secondly, write the number of degrees between the north 
end of the needle and the nearest zero mark ; and thirdly, 
write E. or W. according as the north end of the needle is 
nearer E. or W. of the compass circle. 

In Fig. 5 the bearing would be N. 

In Fig. 6 the bearing would be S. 

In Fig. 7 the bearing would be S. 

In Fig. 8 the bearing would be N. 
If the needle coincides with the N. S. or E.W. line, the bear- 
ing would be N., S., E., or 
W., according as the north 
end of the needle is over 
N., S, E., or W. 

As the compass circle is 
divided into half-degrees, 
the bearing may be deter- 
mined pretty accurately to 
quarter-degrees. 

When a fence or other 
obstruction interferes with 
placing the instrument 
over the line, it may be 

placed at one side, the F( s- 7. Fig. 8. 

flag-staff being placed at an equal distance from the line. 



45° 


W 


45° 


w 


30° 


E. 


60° 


E. 




* The magnetic bearing is meant unless otherwise specified. 



10 



SURVEYING. 



Local Disturbances. Before a bearing is recorded, care should 
be exercised that the chain, pins, and other instruments that 
would affect the direction of the needle, are removed from the 
vicinity of the compass. Even after the greatest care in this 
respect is exercised, the direction of the needle is often affected 
by iron ore, ferruginous rocks, etc. 

Eeverse Bearings. When the bearing of a line has been taken, 
the instrument should be removed to the other end of the line 
and the reverse bearing taken. The number of degrees should 
be the same as before, but the letters should be reversed. 

To take the hearing of a line one end of which cannot he seen 
from the other. Run a random line (§ 4, 3) ; then (Fig. 4), 

tan CAB = ^%; 

whence, the angle CAB may be found. This angle combined 
with the bearing of the random line will give the bearing 
required. 

Another method will be given in § 19. 

To measure a horizontal angle by means of the needle. Place the 
compass over the vertex of the angle, take the bearing of each 
side separately, and combine these bearings. 

To measure angles of elevation. Bring the south end of the 
compass towards you, place the eye at the lower eye-piece, 
and with the hand hold a card on the front side of the north 
sight, so that its top edge will be at right angles to the divided 
edge and coincide with the zero mark ; then, sighting over 
the top of the card, note upon a flag-staff the height cut by 
the line of sight ; move the staff up the elevation, and carry 
the card along the edge of the sight until the line of sight 
again cuts the same height on the staff ; read off the degrees 
of the tangent scale passed over by the card. 

To measure angles of depression. Proceed in the same man- 
ner as above, using the eye-piece and tangent scale on the 
opposite sides of the sights, and reading from the top of the 
sight. 



INSTRUMENTS AND THEIR USES. 



11 



§ 7. Verniers. 



First form. Let AB (Fig. 9) represent a portion of a rod 
for measuring heights (§ 32). The graduation to feet and hun- 
dredths of a foot begins at the lower end, which rests on the 
ground when the rod is in use. The line 
extending nearly across the rod at the bot- 
tom of the portion shown marks the begin- 
ning of the fourth foot. The face of the rod 
is divided into four columns : in the first is 
written the number of feet ; in the second, 
the number of tenths ; and in the third, the 
number of hundredths. 

It is evident that, with the arrangement 
just described, heights could be measured 
only to hundredths of a foot. To enable us 
to find the height more precisely, a contri- 
vance called a Vernier is used. This is shown 
at the right of the rod. It consists of a piece 
of metal or wood, the graduated part of 
which is y 1 -^- of a foot in length ; and this 
is divided into ten equal parts. Hence, one 
division of the vernier = y 1 ^ of -^^ — yxnnj- 
of a foot ; and one division of the vernier 
exceeds one division of the rod by y 11 



rh 



1 (f of a foot. 



Ho" 



side 



Fig. 9. 



The vernier slides along the face 
of the rod. 

To use the vernier, place the lower end 
of the rod upon the ground, and move the vernier until its in- 
dex or zero mark is opposite the point whose distance from 
the ground is desired. In the figure, the height of the index 
of the vernier is evidently 4.16 feet, increased by the distance 
of the index above the next lower line (4.16) of the rod. We 
shall now determine this distance. 



12 



SURVEYING. 



Observe which, line of the vernier is exactly opposite a line 
of the rod. In this case, the line of the vernier numbered 7 is 
opposite a line of the rod. Then, since each division of the 
vernier exceeds each division of the rod by 10 1 00 of a foot, 



6 of the 
5 of the 
4 of the 
3 of the 
2 of the 
1 of the 
of the 

Hence, 

B 



vernier 
vernier 
vernier 
vernier 
vernier 
vernier 
vernier 

the re 



1000 



is ToW of a foot 
of a foot 
of a foot 
of a foot 
of a foot 
of a foot 
of a foot 

readme 



quired 



above the next lower line of the rod. 
above the next lower line of the rod. 
above the next lower line of the rod. 
above the next lower line of the rod. 
above the next lower line of the rod. 
above the next lower line of the rod. 
above the next lower line of the rod. 

; is 4.16 + .007 = 4.167 feet. 





7 
6 
5 
4 
3 
2 
1 












10 

9 
S 

6 

5 
4 
3 
2 
1 



































2 




9 

S 

6 
5 
4 
3 
2 
1 
































































1 






B 

S 
7 
6 
5 
4 
Z 
2 
1 






























































9 













In general, the following rule is evident : 
Move the vernier until its zero line is at the 
required height; read the height to the near- 
est hundredth below the index, and write in 
the thousandth place the number of the divi- 
sion line of the vernier which stands opposite 
any line of the rod. 

Second form. In this form (Fig. 10) the 
graduated part of the vernier is y^-g- of a foot 
in length, and is divided into ten equal parts. 
Hence, one division of the vernier = -^ of 
of a foot ; and one division of the 
than one division of the rod 






100 1000 

vernier is less 



b y ifo- 



of a foot. 



1000 1000 

The height of the index of the vernier 
in Fig. 10 is 4.16 feet, increased by the dis- 
tance of the index from the next lower line 
(4.16) of the rod. We shall now determine 
this distance. 

"We observe that the line of the vernier 
numbered 7 stands exactly opposite a line 
(3) of the rod. Hence, 



INSTRUMENTS AND THEIR USES. 



13 



6 of the vernier is T ^ of a foot above the next lower line of the rod. 
5 of the vernier is T ^ of a foot above the next lower line of the rod. 
4 of the vernier is y^ of a foot above the next lower line of the rod. 
3 of the vernier is j^ of a foot above the next lower line of the rod. 
2 of the vernier is j^ of a foot above the next lower line of the rod. 
1 of the vernier is j^/fo of a foot above the next lower line of the rod. 
of the vernier is T ^ of a foot above the next lower line of the rod. 

Hence, the required reading is 4.16 -f .007 = 4.167 feet ; 
and the rule is evidently the same as for the first form. 




Fig. II. 



Oompass Verniers. Let LL 1 (Fig. 11) represent the limb of 
the compass graduated to half-degrees, and W 1 the vernier 
divided into thirty equal spaces, equal to twenty-nine spaces 
of the limb. Then one space of the vernier is less than one 
space of the limb by 1', and the reading may be obtained to 
single minutes. 

In Fig. 11 the index or zero of the vernier stands between 
32° and 32° 30', and the line of the vernier marked 9 coincides 
with a line of the limb. Hence, the reading is 32° 9'. 

When the index moves from the zero line of the limb in a 
direction contrary to that in which the numbers of the limb 
run, the number of minutes obtained as above must be sub- 
tracted from 30' to obtain the minutes required. 

If, however, the vernier be made double, that is, if it have 
thirty spaces on each side of the zero line, it is always read 
directly. The usual form of the double vernier, shown in 



14 



SURVEYING. 



Fig. 12, has only fifteen spaces on each side of the zero line. 
When the vernier is turned to the right less than 15' past a 
division line of the limb, read the lower figures on the left of 
the zero line at any coincidence ; if moved more than 15' past 
a division line of the limb, read the upper figures on the right 
of the zero line at any coincidence ; and vice versa. 




Uses of the Compass Vernier. The most important use of the 
vernier of the vernier compass is in setting off the variation 
of the needle (§ 23). If 4he variation of the needle at any 
place is known, by means of the vernier screw the compass 
circle may be turned through an arc equal to the variation. 
If the observer stands at the south end of the instrument, the 
vernier is turned to the right or left according as the varia- 
tion is west or east. The compass will now give the bearings 
of the lines with the true meridian. 

In order to retrace the lines of an old survey, turn the sights 
in the direction of a known line, and move the vernier until the 
needle indicates the old bearing. The arc moved over by 
the vernier will indicate the change of variation since the time 
of the old survey. If no line is definitely known, the change 
of variation from the time of the old survey will give the arc 
to be set off. 



INSTRUMENTS AND THEIR USES. 15 



§ 8. The Surveyor's Transit. 

This instrument is shown on page 17. 

The compass circle is similar to that of the compass. The 
vernier plate which carries the telescope has two verniers and 
moves entirely around the graduated limb of the main plate. 
The axis of the telescope carries a vertical circle which meas- 
ures vertical angles to single minutes by means of a vernier. 
Under the telescope, and attached to it, is a spirit level by 
means of which horizontal lines may be run, or the difference 
of level between two stations found. The cross wires are two 
fine fibres of spider's web, or fine platinum wires, which extend 
across the tube of the telescope at right angles to each other ; 
their intersection determines the optical axis or line of colli- 
mation of the telescope. The transit is levelled by four level- 
ling screws which pass through a plate firmly fastened to the 
ball spindle, and the lower ends of which rest in depressions 
on the upper side of the tripod plate. 

Two recent improvements (introduced by the Messrs. Gur- 
ley) enable the surveyor to bring the transit quickly to an 
approximately level position by the pressure of the hands, 
after which the levelling screws are used ; also, to change the 
position of the transit without changing the position of the 
tripod legs, so as to bring the bob exactly over any point. 

To level the transit by the levelling screws. Turn the instru- 
ment until the spirit levels on the vernier plate are parallel 
to the vertical planes passing through opposite pairs of level- 
ling screws. Take hold of opposite screw heads with the 
thumb and fore-finger of each hand, and turn both thumbs 
in or out as may be necessary to raise the lower side of the 
parallel plate and lower the other until the desired correction 
is made. 

To use the telescope. Both the eye-piece and the object 
glass may be moved in and out by a rack-and-pinion move- 
ment. The eye-piece must be moved until the cross wires are 



16 SURVEYING. 



perfectly distinct; in which case, a slight movement of the 
eye of the observer, from side to side, will produce no appar- 
ent change in the position of the threads upon the object. 
The object glass must be moved until the object is distinctly 
visible ; and this operation must be repeated whenever the 
distance of the object is changed. 

§ 9. Uses of the Transit. 

The transit may be used for all the purposes indicated in 
§ 6, but with much greater precision than the compass. The 
principal use, however, of the transit is in measuring horizontal 
angles by means of the graduated limb and verniers. 

To measure a horizontal angle with the transit. Place the transit 
over the vertex of the angle ; level, and set the limb at zero. 
Turn the telescope in the direction of one of the sides of the 
angle, clamp to the spindle ; unclamp the main plate, and turn 
the telescope until it is in the direction of the other side of 
the angle, and read the angle by the verniers. The object 
of the two verniers on the vernier plate is to correct any mis- 
take that might arise from the want of exact coincidence in 
the centres of the verniers and the limb. The correct reading 
may be obtained by adding to the reading of one vernier the 
supplement of the reading of the other, and dividing by 2. 

By turning off a right angle by this method, perpendiculars 
may be constructed with much greater facility than by the 
chain. 

§ 10. The Theodolite. 

The telescope of the transit can perform a complete revo- 
lution on its axis; whence the name transit. The theodolite 
differs from the transit chiefly in that its telescope cannot be so 
revolved. It is not much used in this country. 

§11. The Kailroad Compass. 

This instrument has all the features of the ordinary com- 
pass, and has also a vernier plate and graduated limb for 
measuring horizontal angles. 




THE SURVEYOR'S TRANSIT. 



INSTRUMENTS AND THEIR USES. 



19 



§ 12. Plotting. 

The principal plotting instruments are a ruler, pencil, 
straight-line pen, hair-spring dividers, diagonal scale, a right 
triangle of wood, and a circular protractor. A T-square will 
also be found convenient. 




Fig. 13. 

The Diagonal Scale. A portion of this scale is shown in Fig. 
13. AB is the unit. AB and A'B' are divided into ten 
equal parts, and B is joined with h, the first division point to 
the left of B'; the first division point to the left of B is joined 
with the second to the left of B', etc. 

The part of the horizontal line numbered 1 intercepted be- 
tween BB 1 and Bh is evidently y 1 ^ of tV^tw °f tne unit 5 
the part of the horizontal line numbered 2 intercepted between 
BB 1 and Bh is y^ of the unit, etc. 

The method of using this scale, in laying off distances, will 
be made plain by the following example : 

Let it be required to lay off the distance 1.43. 

Place one foot of the dividers at the intersection of the horizontal line 
numbered 3 and the diagonal numbered 4, and place the other foot at 
the intersection of the vertical line numbered 1 (CC f ) and the horizontal 
line numbered 3 ; the distance between the feet of the dividers will be 
the distance required. For, measuring along the horizontal line num- 
bered 3, from CC to BB' is 1 ; from BB' to Bh is .03 ; and from Bh to 
the diagonal numbered 4 is .4 ; and 1 + .03 + .4 = 1.43. 



20 



SURVEYING. 



The Circular Protractor. This instrument (Fig. 14) usually 
consists of a semi-circular piece of brass or german silver, hav- 
ing its arc divided into degrees and its centre marked. 

To lay off an angle with the protractor, place the centre 
over the vertex of the angle, and make the diameter coincide 
with the given side of the angle. Mark off the number of 
degrees in the given angle, and draw a line through this point 
and the vertex. 




Fig. 14. 

Some protractors have an arm which carries a vernier, by 
which angles may be constructed to single minutes. 

To draw through a given point a line parallel to a given 
line, make one of the sides of a triangle coincide with the 
given line, and, placing a ruler against one of the other sides, 
move the triangle along the ruler until the first side passes 
through the given point ; then draw a line along this side. 

To draw through a given point a line perpendicular to a 
given line, make the hypotenuse of a right triangle coincide 
with the given line, and, placing a ruler against one of the 
other sides of the triangle, revolve the triangle about the ver- 
tex of the right angle as a centre until its other perpendicular 
side is against the ruler ; then move the triangle along the 
ruler until the hypotenuse passes through the given point, 
and draw a line along the hypotenuse. 



CHAPTER II. 
LAND SURVEYING 



§ 13. Definition. 

Land Surveying is the art of measuring, laying out, and 
dividing land, and preparing a plot. 

§ 14. Determination of Areas. 
The unit of land measure is the 

acre = 10 square chains 

— 4 roods 

— 160 square rods, perches, or poles. 

Areas are referred to the horizontal plane, no allowance 
being made for inequalities of surface. 

For convenience of reference, the following rules for areas 
are given: 

Let A, B, and C be the angles of a triangle, and a, b, and c 
the opposite sides, respectively ; and let s = I (a + b -f- c). 

Area of triangle A BC = Vs (s — a) (s — b) (s — c) [a] 

= £ be sin A [b] 

_ ia 2 sini?sin(7 r -, 

~*rin(2H-C) LJ 

= i base X altitude. [d] 

Area of rectangle == base X altitude. 

Area of trapezoid = \ sum of parallel sides X altitude. 

Problem 1. To determine the area of a triangular field. 
Measure the necessary parts with a Gunter's chain, or with a chain 
and transit, and compute by formula [a], [b], [c], or [d]. 



22 



SURVEYING. 



Problem 2. To find the area of a field having any number of 
straight sides. 

(a) Divide the field into triangles by diagonals ; find the area of each 
triangle and take the sum. 

(b) Run a diagonal, and perpendiculars from the opposite vertices to 
this diagonal. The field is thus divided into right triangles, rectangles, 
and trapezoids, the areas of which may be found and the sum taken. 




Fig. 15. 



Fig. 16. 



Problem 3. To find the area of a field having an irregular 
boundary line. 

(a) Let AG BCD (Fig. 15) represent a field having a stream AEFG 
HKB as a boundary line. Run the line AB. From E, F, G, H, and 
K, prominent points on the bank of the stream, let fall perpendiculars 
EE f , FF f , etc., upon AB. Regarding AE, EF, etc., as straight lines, the 
portion of the field cut off by AB is divided into right triangles, rect^ 
angles, and trapezoids, the necessary elements of which can be measured 
and the areas computed. The sum cl these areas added to the area of 
ABCD will give the area required. 

(b) When the irregular boundary line crosses the straight line joining 
its extremities, as in Fig. 16, the areas of AEFH and HGB may be found 
separately, as in the preceding case. Then the area required = ABCD -f 
HGB - AEFH. 



Problem 4. To determine the area of a field from two interior 
stations. 

Let ABCD(Fig. 17) represent a field, and Pand P' two stations within 
it. Measure PP' with great exactness. Measure the angles between PP' 
and the lines from P and P' to the corners of the field. 



DETERMINATION OF AREAS. 



23 



In the triangle PP'D, PP' and the angles P'PD and PP'D are 
known ; hence, PD may be found. In 
like manner, PC may be found. Then, 
in the triangle PD C, PD, PC, and the 
angle DPC are known ; hence, the area 
of PDC may be computed. 

In like manner, the areas of all the 
triangles about P au/ 1 P l may be deter- 
mined. 

Area ABCD = PAD + PDC + PCS 
+ PBA. Also, 

Area ABCD = P'AD + P'DC+P'CB 
+ P'BA. 

Problem 5. To determine the area of a field from two exterior 
stations. d 

Let ABCD (Fig. 18) represent the field, 
and P and P' the stations. Determine the 
areas of the triangles PAD, PDC, PCB, 
and PBA, as in the preceding problem. 

Area ABCD = PAD + PDC + PBC - 
PBA. Also, 

Area ABCD = P' AD + P'DC + P'BA 
- P'BC 





Fig. 18. 



Exercise I. 



1. Required the area of a triangular field whose sides are 
respectively 13, 14, and 15 chains. 

2. Required the area of a triangular field whose sides are 
respectively 20, 30, and 40 chains. 

3. Required the area of a triangular field whose base is 
12.60 chains, and altitude 6.40 chains. 

4. Required the area of a triangular field which has two sides 
4.50 and 3.70 chains, respectively, and the included angle 60°. 

5. Required the area of a field in the form of a trapezium, 
one of whose diagonals is 9 chains, and the two perpendiculars 
upon this diagonal from the opposite vertices 4.50 and 3.25 
chains. 



24 



SURVEYING. 




6. Required the area of the field ABCDEF (Fig. 19), if 
AE= 9.25 chains, FF'= 6.40 chains, PP= 13.75 chains, ED' 
= 7 chains, EB = 10 chains, CC = 
4 chains, and J.J.'=4.75 chains. 

7. Required the area of the field 
ABCEEF (Fig. 20), if 
AF'= 4 chains, FF'= 6 chains, 
EE' = 6.50 chains, A P' = 9 chains, 
AE = 14 chains, A C = 10 chains, 
AB' = 6.50 chains, BB' = 7 chains, 
CC =6.75 chains. 
Required the area of the field AGBCE (Fig. 15), if the 
diagonal AC=d, BB' (the perpen- 
dicular from Bio AC) = 1, EE' (the 
perpendicular from E to AC) = 1.60, 
PP" = 0.25, FF'= 0.25, ££' = 0.60, 
E:E' = 0.d2, XE'=0M,AE'=0.2, 
E'F' = 0.50, P'£' = 0.45, G'H' = 
0.45, fi^K 1 = 0.60, and K'B = 0.40. 
9. Required the area of the field 
AGBCE (Fig. 16), if AB = 3. AC 
= 5, A5=6, angle DAC=4b°, angle £.4(7= 30°, X#' = 
0.75, AF' = 2.25, AM =2.53, AG 1 = 3.15, .££" = 0.60, 
PP' = 0.40, and ££' = 0.75. 

10. Determine the area of the field ABCE from two inte- 
rior stations, P and P', if FP'— 1.50 chains, 

angle PP' C= 89° 35', angle P'PB= 3° 35', 

PP'P = 185° 30', PP.4 = 113° 45', 

PP'J. = 309° 15', P'PP = 165° 40', 

PP'E = 349° 45', F'FC = 303° 15'. 

11. Determine the area of the field ABCE from two exte 
rior stations P and P', if PP' = 1-50 chains, 

angle P'PB= 41° 10', angle PP'P= 66° 45', 

P'P,4 = 55° 45', ~ PP'C= 95° 40', 

P'PC= 77° 20', PP'B = 132° 15', 

P'PE = 104° 45'| PPM = 103° 0'. 




RECTANGULAR SURVEYING. 



25 



KECTANGULAK SURVEYING. 

§ 15. Definitions. 

An East and "West Line is a line perpendicular to the mag- 
netic meridian. 

The Latitude' of a line is the distance between the east and 
west lines through its extremities. 

The Departure of a line is the distance between the meridians 
through its extremities. 

Note. When a line extends north of the initial point the latitude is 
called a northing ; when it extends south, a southing ; when it extends 
^ast the departure is called an easting ; when it extends west, a westing. 

The Meridian Distance of a point is its distance from a meridian. 

Let AB (Fig. 21) represent a line, and NAS the magnetic 
meridian. Let BB' be perpendicular to JVS. 

The bearing of the line AB is the angle 
BAB'. 

The latitude of the line AB is AB'. 

The departure of the line AB is BB'. 

The meridian distance of the point B is 
BB'. 

In the right triangle ABB', 

AB' = ABx cos BAB', 
and BB' = AB x sin BAB'. 

Hence, latitude = distance X cos of bearing, 
and departure = distance X sin of bearing. 

The latitudes and departures correspond- 
ing to any distance and bearing may be 
found from the above formulas by means of 
a table of natural sines and cosines, or from 
" The Traverse Table." * Fig. 21. 



* See Table VII. of Wentworth & Hill's Five-Place Logarithmic and 
Trigonometric Tables. 



26 



SURVEYING. 



§ 16. Field Notes, Computation, and Plotting. 

The field notes are kept in a book provided for the purpose. 
The page is ruled in three columns, in the first of which is 
written the number of the station ; in the second, the bearing 
of the side ; and in the third, the length of the side. 

Example 1. To survey the field ABCD (Fig. 22). 

N 






Field Notes. 


1 


N. 20° E. 


8.66 


2 


S. 70° E. 


5.00 


3 


S. 10° E. 


10.00 


4 


N. 70° W. 


10.00 



(a) To obtain the field notes. 

Place the compass at A, the first sta- 
tion, and take the bearing of AB (§ 6) ; 
suppose it to be N. 20° E. Write the re- 
sult in the second column of the field notes 
opposite the number of the station. Meas- 
ure AB = 8.66 chains, and write the result 
in the third column of the field notes. 

Place the compass at B, and, after test- 
ing the bearing of AB (§ 6), take the 
bearing of BC, measure BC, and write the 

results in the field notes ; and so continue until the bearing and length 

of each side have been recorded. 



Fig. 22. 



(b) To compute 


the area. 














i. 


n. 


m. 


rv. 


V. 


VI. 


VII. 


VIII. 


IX. 


X. 


XI. 


Side. 


Bearing. 


Dist. 


N. 


S. 


E. 


W. 


M.D. 


D. M. D. 


N. A. 


S.A. 


AB 


N. 20° E. 


8.66 


AB' 

8.14 




BB' 
2.96 




BB' 

2.96 


BB' 

2.96 


1ABB' 
24.0944 




BC 


S. 70° E. 


5.00 




B'C 
1.71 


CC" 
4.70 




CC 
7.66 


BB'+CC 

10.62 




2CCBB' 

18.1602 


CD 


S. 10° E. 


10.00 




CD' 

9.85 


DD" 
1.74 




DD' 

9.40 


CC+DD' 

17.06 




ID'DCC 

168.0410 


DA 


N.70°W. 


10.00 


D'A 
3.42 






DD' 

9.40 





DD' 

9.40 


2 ADD' 

32.1480 






56.2424 


186.2012 



/ 

FIELD NOTES. 27 



129.9588 



64.98 sq. chains. 



The survey may begin at any corner of the field ; but in computing 
the area, the field notes should be arranged so lg6 2012 

that the most eastern or most western station 56.2424 

will stand first. For the sake of uniformity, we 2 
shall always begin with the most western station, 10 
and keep the field on the right in passing around it. 6 - 498 acres - 

The field notes occupy the first three of the eleven columns in the 
above tablet. Columns IV., V., VI., and VII. contain the latitudes 
and departures corresponding to the sides, and taken from the Traverse 
Table. The lines represented by these numbers are indicated imme- 
diately above each number. Column VIII. contains the meridian dis- 
tances of the points B, C, D, and A, taken in order. Column IX. contains 
the double meridian distances. Their composition is indicated by the 
letters immediately above the numbers. Column X. contains the pro- 
ducts of the double meridian distances by the northings in the same 
line. The first number, 
24.0944 = 2.96 x 8.14 = BB> x AB' = 2 area of the triangle ABB' • 
32.1480 = 9.40 x 3.42 = DD' X AD* = 2 area of the triangle ADD'. 

Column XI. contains the products of the double meridian distances by 
the southings in the same line. The first number, 

18.1602 = 10.62 x 1.71 = (BB' + CC) x B'C 

= 2 area of the trapezoid C'CBB' ; 
168.0410 = 17.06 x 9.85 = (CC + DD') x D'C 

= 2 area of the trapezoid D'DCQ'. 
The sum of the north areas in column X. 

= 56.2424 = 2(ABB> + ADD?). 
The sum of the south areas in column XI. 

= 186.2012 = 2(CCBB f + D'DCC). 
But (C'CBB' + D'DCC')- (ABB f + ADD') = ABCD. 

Hence, 2(CCBB' + D'DCC) - 2 (ABB' + ADD') = 2 ABCD; 
that is, 186.2012 - 56.2424 = 129.9588 = 2 ABCD. 



Hence, area ABCD = % of 129.9588 = 64.9794 sq. ch. = 6.498 acres. 

(c) To make the plot. 

The plot or map may be drawn to any desired scale. If a line one 
inch in length in the plot represents a line one chain in length, the plot 
is said to be drawn to a scale of one chain to an inch. In this case the 
plot (Fig. 22) is drawn to a scale of eight chains to an inch. 

Draw the line NAS to represent the magnetic meridian, and lay off 
the first northing AB' = 8.14 (§ 12). Draw the indefinite line B'E per- 



28 SURVEYING. 



pendicular to NS and lay off B'B, the first easting = 2.96. Join A and 
B\ then the line AB will represent the first side of the field. Through 
B draw BC" perpendicular to BB', and make BC" = 1.71, the first 
southing. Through C" draw C" C perpendicular to BC", and equal to 
4.70, the second easting. Join B and 0. The line BC will represent 
the second side of the field. 

Proceed in like manner until the field is completely represented. The 
extremity of the last line D 1 A, measured from D' , should fall at A. This 
will be a test of the accuracy of the plot. 

By drawing the diagonal AC, and letting fall upon it perpendiculars 
from B and D, the quadrilateral ABCD is divided into two triangles, 
the bases and altitudes of which may be measured and the area com- 
puted approximately. 

Other methods of plotting will suggest themselves, but the method 
just explained is one of the best. 

Balancing the Work. 

In the survey, we pass entirely around the field ; hence, we 
move just as far north as south. Therefore, the sum of the 
northings should equal the sum of the southings. In like 
manner, the sum of the eastings should equal the sum of the 
westings. In this way the accuracy of the field work may be 
tested. 

In Example 1, the sum of the northings is equal to the sum 
of the southings, being 11.56 in each case ; and the sum of the 
eastings is equal to the sum of the westings, being 9.40 in each 
case. Hence, the work balances. 

In actual practice the work seldom balances. When it does 
not balance, corrections are generally applied to the latitudes 
and departures, by the following rules : 

The perimeter of the field : any one side 

: : total error in latitude : correction ; 
: : total error in departure : correction. 

If special difficulty has been experienced in taking a par- 
ticular bearing, or in measuring a particular line, the correc- 
tions should be applied to the corresponding latitudes and 
departures. 



FIELD NOTES. 



29 



The amount of error allowable varies in the practice of dif- 
ferent surveyors, and according to the nature of the ground. 
An error of 1 link in 8 chains would not be considered too 
great on smooth, level ground; while, on rough ground, an 
error of 1 link in 2 or 3 chains 
might be allowed. If the error 
is considerable, the field meas- 
urements should be repeated. 

Example 2. Let it be re- 
quired to survey the field AB 
CD£F(Fig. 23). 

Field Notes. 



1 


N.73°30'W. 


5.00 


2 


S. 16°30 / W. 


5.00 


3 


N.28°30'W. 


7.07 


4 


N.20°00'E. 


11.18 


5 


S. 43° 3C E. 


5.00 


6 


S. 13°3(XE. 


10.00 



81.4955 



161.5933 



).7967 



8.0797 acres. 



Explanation. The first station 
in the field notes is D, but we re- 
arrange the numbers in the tablet so 
that A stands first. The northings 
and southings balance, but the east- 
ings exceed the westings by 1 link. 
We apply the correction to the west- 
ing 4.79 (the distance DE being in 
doubt), making it 4.80, and rewrite 

all the latitudes and departures in the next four columns, incorporating 
the correction. In practice, the corrected numbers are written in red ink. 







* fej b 9 to su 

^ ^1 fe| b C5 bs 


Co 






!zj QD fej m w >zj 

Jo M ^ l-l rf*. to 

oo © co co co © 

co co co co co o 
© o o © © © 

4 4 4 w « h 


CD 






11.18 
5.00 

10.00 
5.00 
5.00 
7.07 


5 

5" 




00 


OS % M * * © 

to * k- ! ! t" 


«; 




00 


• *• '. » to | 
" ^ ', ^i b " 

© to CO 


P> 




CO 
© 


tO CO CO 

co *>. bo 

CO 1^ tO 


fi 




JO 

CO 


CO M *» 

W I|i M ' 

-* to to 


^ 






AB' 
10.51 

D'E' 
1.42 

F'A 
6.21 


^ 






** tq • ©O cofca . 

ioS : Sb l§q : 


Co 






B'B 

3.82 
C"C 
3.44 
D"D 
2.33 


1] 






co^sj mS-j *.b ; ; ; 
co i *>- : bo o . 


^. 






f ^ ^feq j°b r>C5 phs 

® co,--ji-oiLjk>i--oo,' 

-irq ©£q ©O o>0 ^03 








B'B 

3.82 

B'B+C'C 

11.08 
C'C+D'D 

16.85 
D'D + E'E 

14.38 

E'E+ F'F 

8.16 

F'F 

3.37 


CD 




CO 

o 




to 

kb 




^to 
Pk 

8* 


lb. 




CO 

© 
co 




tO 

CO*J 


2CCBB' 

40.2204 
ID'DCC 

163.7820 







30 



SURVEYING. 



The remainder of the computation does not require expla- 
nation. 

It will be seen that this method of computing areas is 
perfectly general. 

# §17. Supplying Omissions. 

If, for any reason, the bearing 
and length of any side do not ap- 
pear in the field notes, the latitude 
and departure of this side may be 
found in the following manner : 

Find the latitudes and departures 
of the other sides as usual. The 
difference between the northings 
and southings will give the north- 
ing or southing of the unknown 
side, and the difference between 
the eastings and westings will give 
the easting or westing of the un- 
known side. 

If the length and bearing of the 
unknown side are desired, they 
may be found by solving the right 
triangle, whose sides are the lati- 
tude and departure found by the method just explained, and 
whose hypotenuse is the length required. 




Fig. 23 



§ 18. Irregular Boundaries. 

If a field have irregular boundaries, its area may be found 
by offsets, as explained in § 14, Prob. 3.- 



§ 19. Obstructions. 

If the end of a line be not visible from its beginning, or if 
the line be inaccessible, its length and bearing may be found 
as follows : 



OBSTRUCTIONS. 



31 



1. By means of a random line (§ 4, 3). 

2. When it is impossible to run a random line, which is 
frequently the case on account of the extent of the obstruction, 
the following method may be used : N 

Let AB (Fig. 24) represent an inaccessible line 
whose extremities A and B only are known, and 
B invisible from A. 

Set flag-staffs at convenient points, C and D. 
Find the bearings and lengths of A C, CD, and DB, 
and then proceed to find the latitude and depar- 
ture of AB. as in § 17. 



Fig. 24. 



Example. Suppose that we have the following notes (see 
Fig. 24): 



Side. 


Bearing. 


Dlst. 


N. 


S. 


E r - 


- w. 


AC 
CD 
DB 


S. 45° E. 

E. 

N.30°E. 


3.00 
3.50 
4.83 


4.18 


2.12 


2.12 
3.50 
2.42 






4.18 


2.12 


8.04 







Fig. 25. 



4 is The northing of AB is 2.06, and the easting, 8.04 ; which 

2 12 numbers may be entered in the tablet in the columns N. and E., 
TI7 opposite the side AB. 

If the bearing and length of AB are required, construct the 
right triangle ABC (Fig. 25), making AC= 8.04 and BC= 2.06. 
, nAn BC 2.06 

Hence, the angle BAC = 14° 22'. 



0.256. 



Also, AB =VAC 2 + 'BC 1 = V8.04 2 + 2.06 2 = 8.29. 

Therefore, the bearing and length of AB are N. 75° 38' E. 8.29. 



32 



SURVEYING. 



Exercise II. 

Note. In examples 5 and 6 detours were made 
sible sides (§ 19, 2). The notes of the detours are 
1. 5. 



on account of inacces- 
written in braces. 
8. 



Sta. 


Bearings. 


Diet. 


1 


S. 75° E. 


6.00 


2 


S. 15° E. 


4.00 


3 


S. 75° W. 


6.93 


4 


N.45°E. 


5.00 


5 


N.45°W. 


5.19J 





2. 




Sta. 


Bearings. 


Dist. 


1 


N.45°E. 


10.00 


2 


S. 75° E. 


11.55 


3 


S. 15° W. 


18.21 


4 


N.45°W. 


19.11 





3. 




Sta. 


Bearings. 


Dist. 


1 


N.15°E. 


3.00 


2 


N.75°E. 


6.00 


3 


S. 15° W. 


6.00 


4 


N.75°W. 


5.20 



4. 



Sta. 


Bearings. 


Dist. 


1 


N.89°45'E. 


4.94 


2 


S. 7°0O / W. 


2.30 


3 


S. 28°0O / E. 


1.52 


4 


S. 0°45'E. 


2.57 


5 


N.84°45'W. 


5.11 


6 


N. 2°30 / W. 


5.79 



Sta. 
1 


Bearings. 


Dist. 


S. 2°15 / E. 


9.68 


f 


N.51°45 / W. 


2.39 


2 


S. 85°0O'W. 


6.47 


I 


S. 55°10 / W. 


1.62 


3 


N. 3°45 / E. 


6.39 


4 


S. 66°46'E. 


1.70 


5 


N.15°00'E. 


4.98 


6 


S. 82°45 / E. 


6.03 



Sta. Bearings, 


Dist. 


■{ 


S. 81°20 / W. 


4.28 


N.76°3(yW. 


2.67 


2 


N. 5°(XyE. 


8.68 


3 


S. 87°30 / E. 


5.54 




S. 7°0O / E. 


1.79 


4- 


S. 27°0O / E. 
S. 10°3(/E. 


1.94 
5.35 


' 


N.76°45 / W. 


1.70 



Sta. 


Bearings. 


Dist. 


1 


N. 6°15'W. 


6.31 


2 


S. 81°5(yW. 


4.06 


3 


S. 5°0O'E. 


5.86 


4 


N.88°3(/E. 


4.12 



Sfa. 
1 


Bearings. 


Dist. 


N. 5°3(yW. 


6.08 


2 


S. 82°3(yW. 


6.51 


3 


S. 3°0O / E. 


5.33 


4 


E. 


6.72 





9. 




Sta. 


Bearings. 


Dist. 


1 


N.20°00'E. 


4.62^ 


2 


N.73°0O / E. 


4.16J 


3 


S. 45°15'E. 


6.18J 


4 


S. 38°30 / W. 


8.00 


5 


Wanting. 


Wanting. 



10. 



Sta. 


Bearings. 


£>/»*. 


1 


S. 3°0O / E. 


4.23 


2 


S. 86045^. 


4.78 


3 


S. 37°0O / W. 


2.00 


4 


N.81°0(yW. 


7.45 


5 


N.61°0O / W. 


2.17 


6 


N.32°0O'E. 


8.68 


7 


S. 75°50 / E. 


6.38 


8 


S. 14°45 / W. 


0.98 


9 


S. 79°15 , E. 


4.52 




rectangular method. 33 

§ 20. Modification of the Rectangular Method. 

The area of a field may be found by a modification of the 
rectangular method, if its sides and interior angles are known. 

Let A, B, C, D, represent the inte- 
rior angles of the field A BCD (Fig. 
26). Let the side AB determine the 
direction of reference. 

The bearing of AB, with reference 
to AB, is 0°. 

The bearing of BC, with reference 
to AB, is the angle b = 180°- .#. 

The bearing of CD, with reference 
to AB, is the angle c~C~b. F 'g- 26 - 

The bearing of DA, with reference to AB, is the angle d= A. 

The area may now be computed by the rectangular method, 
regarding AB as the magnetic meridian. 

As the interior angles may be measured with considerable 
accuracy by the transit, the latitudes and departures should 
be obtained by using a table of natural sines and cosines. 

Exercise III. 

1. Find the area of the field ABCD, in which the angle 
.4 = 120°, ^ = 60°, (7=150°, and D = 30°; and the side 
AB = 4 chains, BC— 4 chains, CD = 6.928 chains, and DA 
= 8 chains. 

2. Find the area of the farm ABCDE, in which the angle 
A = 106° 19', B = 99° 40', C= 120° 20', D = 86° 8', and E= 
127°33'; and the side AB= 79.86 rods, £0=121.13 rods, 
CD = 90 rods, DE= 100.65 rods, and EA = 100 rods. 

§21. General Remarks on Determining Areas. 

Operations depending upon the reading of the magnetic 
needle must lack accuracy. Hence, when great accuracy is 
required (which is seldom the case in land surveying), the 
rectangular method (§§ 16-19) cannot be employed. 



34 SURVEYING. 



The best results are obtained by the methods explained in 
§§ 14 and 20, the horizontal angles being measured with the 
transit, and great care exercised in measuring the lines. 

§22. The Variation of the Needle. 

The Magnetic Decimation, or variation of the needle, at any- 
place, is the angle which the magnetic meridian makes with 
the true meridian, or north and south line. The variation is 
said to be east or west, according as the north end of the 
needle lies east or west of the true meridian. Western vari- 
ation is indicated by the sign -j-, and eastern variation by the 
sign -. 

Irregular Variations are sudden deflections of the needle, 
which occur without apparent cause. They are sometimes 
accompanied by auroral displays and thunder storms, and are 
most frequent in years of greatest sun-spot activity. 

Solar-Diurnal Variation. North of the equator, the north end 
of the needle moves to the west, from 8 A.M. to 1.30 p.m., about 
6' in winter and 11' in summer, and then returns gradually to 
its normal position. 

Secular Variation is a change in the same direction for about 
a century and a half; then in the opposite direction for about 
the same time. 

The line of no variation, or the Agonic Line, is a line joining 
those places at which the magnetic meridian coincides with 
the true meridian. In the United States, this line at present 
(1882) passes through Michigan, Ohio, the Virginias, and the 
Carolinas. It is moving gradually westward, so that the varia- 
tion is increasing at places east of this line, and decreasing at 
places west of this line. East of this line the variation is 
westerly, and west of this line the variation is easterly. 

The table on page 35, which has been prepared by permis- 
sion from the report for 1879 of the United States Coast and 
Geodetic Survey, shows the magnetic variation at different 
places in the United States and Canada for several years; 
also, the annual change for 1880. 



VARIATION OF THE NEEDLE. 



35 




O 8» 



o y- 






Po^o <Lo^ro £T2 fo -° -'- 






kJP 



o 



sST^^i 



p 
o-~ o ^ 

QOQ ^ £ - 

P 



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36 



SURVEYING. 



§23. To Establish a True Meridian. 

This may be done as follows : 

1. By means of Burt's Solar Compass (§ 25). 

2. By observations of Polaris. 

The North Star or Polaris revolves about the pole at pres- 
ent at the distance of about 1\° ; hence, it is on the meridian 
twice in 23 h. 56 m. (a siderial day), once above the pole 
called the upper culmination, and 11 h. 58 m. later below the 
pole called the lower culmination. It attains its greatest 
eastern or western elongation, or greatest distance from the 
meridian, 5 h. 59 m. after the culmination. 

The following table gives the time of the upper culmination 
of Polaris for 1882. The time is growing later at the rate of 
about one minute in three years. 



Month. 


First Day. 


Eleventh Day. 


Twenty-first Day. 




H, M. 


H. M. 




H. M. 


January . . 


6 32 p.m. 


5 53 


P.M. 


5 13 p.m. 


February . 


4 30 p.m. 


3 50 


P.M. 


3 11 P.M. 


March . . . 


2 35 p.m. 


1 56 


P.M. 


1 17 P.M. 


April. . . . 


12 29 p.m. 


11 50 


A.M. 


11 11 A.M. 


May .... 


10 31 A.M. 


9 52 


A.M. 


9 13 A.M. 


June .... 


8 30 a.m. 


7 51 


A.M. 


7 12 A.M. 


July 


6 32 a.m. 


5 53 


A.M. 


5 14 A.M. 


August. . . 


4 31 A.M. 


3 52 


A.M. 


3 13 A.M. 


September . 


2 29 a.m. 


1 50 


A.M. 


1 11 A.M. 


October . . 


12 32 a.m. 


11 52 


P.M. 


11 13 P.M. 


November . 


10 30 p.m. 


9 50 


P.M. 


9 11 P.M. 


December . 


8 32 p.m. 


7 52 


P.M. 


7 13 p.m. 



The time of the upper culmination of Polaris may be found 
by means of the star Alioth, which is the star in the handle 
of the Dipper (in the constellation of the Great Bear) next to 
the four which form the bowl. It crosses the meridian about 
the same time as Polaris. Suspend from a height of about 20 
feet a plumb-line, placing the bob in a pail of water to lessen 



TO ESTABLISH A TRUE MERIDIAN. 



37 



its vibrations. About 15 feet south, of the plumb-line, upon a 
horizontal board firmly supported at a convenient height, place 
a compass sight fastened to a board a few inches square. At 
night, when Alioth by estimation approaches the meridian, 
place the compass sight in line with Polaris and the plumb- 
line, and move it so 



as to keep it in this 
line until the plumb- 
line also falls on Ali- 
oth (Fig 27). Note 
the time ; then twen- 
ty-two minutes later 
Polaris will be on the 
meridian. 

If the lower cul- 
mination takes place 
at night,the time may 
be found in a similar 



Pole. 



Pole. 



Fig. 27. 



manner. 

Instead of the compass sight, any upright with a small open- 
ing or slit may be used. 

The plumb-line may be made visible by a light held near it. 

(a) To locate the true meridian by the position of Polaris at its 
culmination. 

1. By using the apparatus described in finding the time of 
culmination. At the time of culmination bring Polaris, the 
plumb-line, and the compass sight into line. The compass 
sight and the plumb-bob will give two points in the true 
meridian. 

2. By means of the transit. Bring the telescope to bear on 
Polaris at the time of culmination, holding a light near to 
make the wires visible, if the observation is made at night. 
The telescope will then lie in the plane of the meridian, which 
may be marked by bringing the telescope to a horizontal 
position. 



38 



SURVEYING. 



(b) To locate the meridian by the position of Polaris at greatest 
elongation. 

The Azimuth of a star is the angle which the meridian plane 
makes with a vertical plane passing through the star and the 
zenith of observer. 

Let Z (Fig. 28) represent the zenith of the place, P the pole, and N 
Polaris at its greatest elongation. Let ZP, ZN, and PN be 
arcs of great circles, and let N be a right angle. 

sin PN= cos (90° - £P)cos(90° - Z). 

[Spher. Trig, g 47-] 
But ZP = the complement of the latitude. 

Hence, 90° — ZP = the latitude of the place. 
Hence, sin PN= cos latitude X sin Z. 
sin PN 



Hence, 



smZ ■ 



cos latitude 



Hence, Z (the azimuth of Polaris) can be found if the lati- 
tude of the place and the greatest elongation of Polaris {PN) 
are known. 

The following table gives the latter element Jan. 1, 1882-93. 

Greatest Elongation of Polaris. 



1882 


1° 19' 12.5" 


1886 


1° 17' 56.6" 


1890 


1° 16' 40.7" 


1883 


1° 18' 53.5" 


1887 


1° 17' 37.6" 


1891 


1° 16' 21.8" 


1884 


1° 18' 34.5" 


1888 


1° 17' 18.6" 


1892 


1°16' 3" 


1885 


1° 18' 15.5" 


1889 


1° 16' 59.7" 


1893 


1° 15' 44.1" 



The greatest elongation of Polaris, or the polar distance, is 
given in the Nautical Almanac. The table gives this element 
for Jan. 1. It may be found for other dates by interpolation. 

To obtain a line in the direction of Polaris at greatest elongation. 

1. By using the apparatus for finding the time of culmina- 
tion. A few minutes before the time of greatest elongation 
(5 h. 59 m. after culmination), place the compass sight in line 
with the plumb-line and Polaris, and keep it in line with these 
until the star begins to recede. At this moment the sight and 
plumb-line are in the required line. 



DIVIDING LAND. 



39 



2. By means of the transit. A few minutes before the time 
of greatest elongation, bring the telescope to bear on the star, 
and follow it, keeping the vertical wire over the star until it 
begins to recede. The telescope will then be in the required 
line. 

To establish the meridian. Having the transit sighted in the 
direction of the line just found, turn it through an angle equal 
to the azimuth in the proper direction. 



§ 24. Dividing Land. 

A few problems in the Division of Land are solved in this 
section, for the purpose of acquainting the student with some 
of the methods employed. The surveyor must, for the most 
part, depend on his general knowledge of Geometry and Trigo- 
nometry, and his own ingenuity, for solutions of the problems 
that arise in practice. 

Problem 1. To divide a triangular field into two parts having a 
given ratio, by a line through a given vertex. A 

Let ABC (Fig. 29) be the triangle, and A the 
given vertex. 

Divide BC at D, so that — — equals the given 

ratio, and join A and D. ABJD and ADC wi 
be the parts required ; for 

ABD : ADC : : BD : DC. 

Problem 2. To cut off from a 
triangular field a given area, by a 
line parallel to the base. 

Let ABC (Fig. 30) be the triangle, 
and let DE be the division line re- 
quired. 

VABC : VADE ::AB: AD. 



AD 



= ABR- 



ADE 
ABC 




40 



SURVEYING. 



Problem 3. To divide any field into two parts having a given 
ratio, by a line through a given point in 
the perimeter. 

Let ABODE (Fig. 31) represent the field, 
P the given point, and PQ the required divi- 
sion line. 

The areas of the whole field and of the 
required parts having been determined, run 
the line PD from P to a corner D, dividing 
the field, as near as possible, as required. 
Determine the area PBCD. 

The triangle PDQ represents the part 
which must be added to PBCD to make the 
required division. 

IxPDxDQxsmPDQ. 

2 area PDQ 




Fig. 31. 



Area PDQ 
Hence, DQ 



Note. DQ = 



PDx sin PDQ 
2 area PDQ 



perpendicular from P on DE 
P on DE may be run and measured directly. 



This perpendicular from 



Problem 4. To divide a field into a given number of parts, 
b so that access to a pond of water is given 

to each. 

Let ABODE (Fig. 32) represent the field, 
and Pthe pond. Let it be required to divide 
the field into four parts. Find the area of 
the field and of each part. 

Let AP be one division line. Eun PE, 
and find the area APE. Take the differ- 
ence between APE and the area of one of 
the required parts ; this will give the area 
of the triangle PQE, from which QE may be 
D found, as in Problem 3. Join P and Q ; 

Flg - 32 - PAQ will be one of the required parts. In 

like manner, PQR and PAS are determined ; whence, PSR must be the 
fourth part required. 




EXAMPLES. 41 



Exercise IV. 

1. From the square ABCD, containing 6 i 1 E, 24 p., 
part off 3 A. by a line EF parallel to AB. 

2. From the rectangle ABCI), containing 8 A. 1 B. 24 p., 
part off 2 A. 1 b. 32 p. by a line EF parallel to AD = 7 ch. 
Then, from the remainder of the rectangle, part off 2 A. 3 B. 
25 p., by a line GH parallel to EB. 

3. Part off 6 A. 3 b. 12 p. from a rectangle ABCD, con- 
taining 15 A., by a line EF parallel to AB ; AD being 10 ch. 

4. From a square ABCD, whose side is 9 ch., part off a 
triangle which shall contain 2 A. 1 B. 36 p., by a line BE 
drawn from B to the side AD. 

5. From ABCD, representing a rectangle, whose length is 
12.65 ch., and breadth 7.58 ch., part off a trapezoid which 
shall contain 7 A. 3 b. 24 p., by a line BE drawn from B to 
the side DC. 

6. In the triangle ABC, AB = 12 ch., AC= 10 ch., and 
BC=8 ch. ; part off 1 A. 2 b. 16 p., by the line DE parallel 
to AB. 

7. In the triangle ABC, AB = 26 ch., AC=20 ch., and 
BC= 16 ch. ; part off 6 A. 1 B. 24 p., by the line DE par- 
allel to AB. 

8. It is required to divide the triangular field ABC among 
three persons whose claims are as the numbers 2, 3, and 5, so 
that they may all have the use of a watering-place at C\ AB 
= 10 ch., AO= 6.85 ch., and CB = 6.10 ch. 

9. Divide the five-sided field ABCHE among three per- 
sons, X, Y, and Z, in proportion to their claims, X paying 
$500, Y paying $750, and Z paying $1000, so that each may 
have the use of an interior pond at P, the quality of the land 
being equal throughout. Given AB = 8.64 ch., BC= 8.27 ch., 
CH= 8.06 ch, HE= 6.82 ch, and EA « 9.90 ch. The per- 
pendicular PD upon AB = 5.60 ch, PD'upon BC= 6.08 ch, 
PD" upon 03"= 4.80 ch, PD'" upon MU=5M ch, and 
PD"" upon EA = 5.40 ch. Assume PIT as the divisional 



42 SURVEYING. 



fence between X 's and Z 's shares ; it is required to determine 
the position of the fences PM and PN between X 's and Y 's 
shares and Y 's and Z 's shares, respectively. 

10. Divide the triangular field ABC, whose sides AB, AC, 
and BC are 15, 12, and 10 eh., respectively, into three equal 
parts, by fences EO and DF parallel to BC. 

11. Divide the triangular field ABC, whose sides AB, BC, 
and A C are 22, 17, and 15 ch., respectively, among three per- 
sons, A, B, and 0, by fences parallel to the base AB, so that 
A may have 3 A., B, 4 A., and 0, the remainder. 



25. United States Public Lands. 
Burt's Solar Compass. 

This instrument, which is exhibited on the following page, 
may be used for most of the purposes of a compass or transit. 
Its most important use, however, is to run north and south 
lines in laying out the public lands. 

A full description of the solar compass, with its principles, 
adjustments, and uses, forms the subject of a considerable vol- 
ume, which should be in the hands of the surveyor who uses 
this instrument. The limits of our space will allow only a 
brief reference to its principal features. 

The main plate and standards resemble these parts of the 
compass. 

a is the latitude arc. 

b is the declination arc. 

h is an arm, on each end of which is a solar lens having its 
focus on a silvered plate on the other end. 

c is the hour arc. 

n is the needle-box, which has an arc of about 36°. 

To run a north and south line with the solar compass. Set off 

the declination of the sun on the declination arc. Set off the 
latitude of the place (which may be determined by this instru- 
ment) on the latitude arc. Set the instrument over the station, 




BURT'S SOLAR COMPASS. 



LAYING OUT PUBLIC LANDS. 



45 



level, and turn the sights in a north and south direction, ap- 
proximately, by the needle. Turn the solar lens toward the 
sun, and bring the su-n's image between the equatorial lines on 
the silvered plate. Allowance being made for refraction, the 
sights will then indicate a true north and south line. 

The Solar Attachment for Transits. 

This consists, essentially, of the solar apparatus of the solar 
compass attached to the telescope of the ordinary transit. 



Laying Out the Public Lands. 

The public lands north of the Ohio River and west of the 
Mississippi are generally laid out in townships approximately 
six miles square. 

A Principal Meridian, or true north and south line, is first 
run by means of Burt's Solar Compass, and then an east and 
west line, called a Base Line. 

Parallels to the base line are run at intervals of six miles, 
and north and south lines at 
the same intervals. Thus, the 
tract would be divided into 
townships exactly six miles 
square, if it were not for the 
convergence of the meridians 
on account of the curvature of w ~~ 
the earth. 

The north and south lines, 
or meridians, are called Eange 
Lines. The east and west lines, 
or parallels, are called Town- 
ship Lines. 

Let jOT (Fig. 33) represent a principal meridian, and WE 
a base line ; and let the other lines represent meridians and 
parallels at intervals of six miles. 

The small squares, A, L, C, etc., will represent townships. 



s 

Fig. 33. 



46 



SURVEYING. 



6 


5 


4 


3 


2 


1 


7 


8 


9 


10 


11 


12 


18 


17 


16 


15 


14 


13 


19 


20 


21 


22 


23 


24 


30 


29 


28 


27 


26 


25 


31 


32 


33 


34 


35 


36 



Fig. 34. 

asN.E., N.W.. S.E 



A would be designated thus : T. 3N..R2W.; that is, 
township three north, range two west ; which means that the 
township is in the third tier north of the base line, and in the 
second tier west of the principal meridian. B and C, respec- 
tively, would be designated thus : T. 4 S., R 3 W. ; and 
T. 2 N., R 2 E. 

The townships are divided into sections approximately one 
mile square, and the sections are di- 
vided into quarter-sections. The town- 
ship, section, and quarter-section corners 
are permanently marked. 

The sections are numbered, beginning 
at the north-east corner, as in Fig. 34, 
which represents a township divided 
into sections. The quarter-sections are 
designated, according to their position, 
. and S.W. 

Every fifth parallel is called a Standard Parallel or Correction 
Line. 

Let JVS (Fig. 35) represent a principal meridian; WE a base 

line; op, etc., meridians; and 
ms the fifth parallel. If Op 
equals six miles, mr will be 
less than six miles on account 
of the convergence of the 
meridians. Surveyors are 
instructed to make Op such 
a distance that mr shall be 
six miles; then mh, hk, etc., 
are taken similarly. At the 
correction lines north of ms 
the same operation is re- 
peated. 

The township and section 
lines are surveyed in such an 
order as to throw the errors on the north and outer townships 
and sections. 





\ 


» .' 1* - 1 





































p 



Fig. 35. 



PLANE-TABLE SURVEYING. 47 

If, in running a line, a navigable stream or a lake more than 
one mile in length is encountered, it is meandered by marking 
the intersection of the line with the bank and running lines 
from this point along the bank to prominent points which are 
marked, and the lengths and bearings of the connecting lines 
recorded. 

Six principal meridians have been established and con- 
nected. In addition to these there are several independent 
meridians in the Western States and Territories which will in 
time be connected with each other and with the eastern 
system. 

§ 26. Plane-Table Surveying* 

After the principal lines of a survey have been determined 
and plotted, the details of the plot may be filled in by means 
of the plane-table ; or, when a plot only of a tract of land is 
desired, this instrument affords the most expeditious means of 
obtaining it. 

An approved form of the plane-table, as used in the United 
States Coast and Geodetic Survey, is shown in the plate on 
page 49. 

The Table-top is a board of well-seasoned wood, panelled 
with the grain at right angles to prevent warping, and sup- 
ported at a convenient height by a Tripod and Levelling 
Head. 

The Alidade is a ruler of brass or steel supporting a telescope 
or sight standards, whose line of sight is parallel to a plane 
perpendicular to the lower side of the ruler, and embracing 
its fiducial edge. 

The Declinatoire consists of a detached rectangular box con- 
taining a magnetic needle which moves over an arc of about 
10° on each side of the 0. 



*In preparing this section the writer has consulted, hy permission, the treat- 
ise on the plane-table hy Mr. E. Hergesheimer, contained in the report for 1880 
of the U.S. Coast and Geodetie Survey. 



48 



SURVEYING. 



Two spirit levels at right angles are attached to the ruler or 
to the declinatoire. In some instruments these are replaced 
by a circular level, or by a detached spirit level. 

The paper upon which the plot is to be made or completed 
is fastened evenly to the board by clamps, the surplus paper 
being loosely rolled under the sides of the board. 

To place the table in position. This operation, which is some- 
times called orienting the table, consists in placing the table 
so that the lines of the plot shall be parallel to the correspond- 
ing lines on the ground. 

This may be accomplished by turning the table until the 
needle of the declinatoire indicates the same bearing as at a 
previous station, the edge of the declinatoire coinciding with 
the same line on the paper at both stations. 

If, however, the line connecting the station at which the 
instrument is placed with another determined station is already 
plotted, the table may be placed in position by placing it over 
the station so that the plotted line is by estimation over and 
in the direction of the line on the ground ; then making the 
edge of the ruler coincide with the plotted line, and turning 
the board until the line of sight bisects the signal at the other 
end of the line on the ground. 

To plot any point. Let ab on the paper represent the line 
AB on the ground ; it is required to plot c, representing C on 
the ground. 

1. By intersection. 

Place the table in position at A (Fig. 36), plumbing a over A, and 
c making the fiducial edge of the 

ruler pass through a; turn the 
alidade about a until the line of 
sight bisects the signal at C, and 
draw a line along the fiducial edge 
of the ruler. Place the table in 
position at B, plumbing b over B, 
and repeat the operation just de- 
scribed, c will be the intersec- 
ts- 36 - tion of the two lines thus drawn. 



+ 






/ N > 




/ X 






/ 


N. 










>1 




s 




x v 


























i)i i.3 
















THE PLANE-TABLE. 



PLANE-TABLE SUEVEYING. 



51 



2. By resection. 

Place the table in position at A (Fig. 37), and draw a line in the di- 
rection of C, as in the former case; then remove the instrument to C, 
place it in position by the line drawn 



X c 















/ 














■^ 






/ 




..' 


b ^a 


n B 



Fig. 37. 



from a, make the edge of the ruler 
pass through b, and turn the alidade 
about b until B is in the line of 
sight. A line drawn along the 
edge of the ruler will intersect the 
line from a in c. 

3. By radiation. 

Place the table in position at A 
(Fig. 38), and draw a line from a 
toward C, as in the former cases. 
Measure AC, and lay off ac to the 
same scale as ab. 

To plot a field ABCD 

1. By radiation. 
Set up the table at any point P, 

and mark p on the paper over P. 6 
Draw indefinite lines from p to- 
ward A, B, C, Measure PA, 

PB , and lay off pa, pb, , to a suitable scale, and join a and b, 

b and c, c and d, etc. 

2. By progression. 

Set up the table at A, and draw a line from a toward B. Measure 
AB, and plot ab to a suitable scale. Set up the table in position at B, 
and in like manner determine and plot be, etc. 

3. By intersection. 

Plot one side as a base line. Plot the other corners by the method of 
intersection, and join. 

4. By resection. 

Plot one side as a base line. Plot the other corners by the method of 
resection, and join. 




Fig. 38. 



52 



SURVEYING. 



The Three Point Problem. 

Let A, B, C represent three points determined and plotted 
as a, b, c (Fig. 39) ; it is required to plot by the plane-table, 
d representing a fourth undetermined point D. 




Fig. 39. 

Determine the point d by the method of resection ; first, 
from A and JB, then from A and C. If the three lines from 
a, b, and c meet in a point, this will determine d. Ordinarily, 
they will not meet in a point, but will form a triangle of error, 
ac be ab. Through a, b, and ab ; a, c, and ac ; and b, c, and 
be, respectively, pass circumferences of circles ; these three 
circles will intersect in the required point d. 






CHAPTER III. 
TRIANGULATION.* 




§27. Introductory Remarks. 

Geographical positions upon the surface of the earth arc 
commonly determined by systems of triangles which connect a 
carefully determined base line with the points to be located. 

Let F (Fig. 40) represent a point whose position with refer- 
ence to the base line AB is 
required. Connect AB with 
F by the series of triangles 
ABC, ACB, ADF, and 
BFF, so that a signal at C 
is visible from A and B, a 
signal at D visible from A 
and C, a signal at E visible 
from A and B, and a signal at F visible from D and F. In 
the triangle ABC, the side AB is known, and the angles at 
A and B may be measured; hence, AC may be computed. 
In the triangle ACB, AC is known, and the angles at A and 
C may be measured ; hence, AB may be computed. In like 
manner BF and FF or DF may be determined. BF, or 
some suitable line connected with BF, may be measured, and 
this result compared with the computed value to test the 
accuracy of the field measurements. 



* In preparing this chapter the writer has consulted, by permission, recent 
reports of the United States Coast and Geodetic Survey. 



Fig. 40. 



54 SURVEYING. 



Three orders of triangulation are recognized ; viz. : 

Primary, in which the sides are from 20 to 150 miles in 
length. 

Secondary, in which the sides are from 5 to 40 miles in 
length, and which connect the primary with the tertiary. 

Tertiary, in which the sides are seldom over 5 miles in 
length, and which bring the survey down to such dimensions 
as to admit of the minor details being filled in by the compass 
and plane-table. 

§ 28. The Measurement of Base Lines. 

Base lines should be measured with a degree of accuracy 
corresponding to their importance. 

Suitable ground must be selected and cleared of all obstruc- 
tions. Each extremity of the line may be marked by cross 
lines on the head of a copper tack driven into a stub which is 
sunk to the surface of the ground. Poles are set up in line 
about half a mile apart, the alignment being controlled by a 
transit placed over one end of the line. 

The preliminary measurement may be made with an iron 
wire about one-eighth of an inch in diameter and 60 m in 
length. In measuring, the wire is brought into line by means 
of a transit set up in line not more than one-fourth of a mile 
in the rear. The end of each 60 m is marked with pencil lines 
on a wooden bench whose legs are thrust into the ground after 
its position has been approximately determined. If the last 
measurement exceeds or falls short of the extremity of the line, 
the difference may be measured with the 20 m chain. 

The final measurement is made with the base apparatus, 
which consists of bars 6 m long, which are supported upon 
trestles when in use. These bars are placed end to end, and 
brought to a horizontal position, if this can be quickly accom- 
plished ; if not, the angle of inclination is taken by a sector, 
or a vertical offset is measured with the aid of a transit, so 
that the exact horizontal distance can be computed. 



MEASUREMENT OF ANGLES. 55 

A thermometer is attached to each bar, so that the tempera- 
ture of the bar may be noted and a correction for temperature 
applied. 

The method of measuring lines varies according to the re- 
quired degree of accuracy in any particular case, but the brief 
description given above will give the student a general idea 
of the methods employed. 

§ 29. The Measurement of Angles. 

Angles are measured by large theodolites, and the reading 
taken by microscopes to single seconds. 

In order to eliminate errors of observation, and errors aris- 
ing from imperfect graduation of the circles, a large number 
of readings is made and their mean taken. Two methods are 
in use; viz., repetition and series. 

The method of repetition consists, essentially, in measuring 
the angles about a point singly, then taking two adjacent 
angles as a single angle, then three, etc.; thus "closing the 
horizon," or measuring the whole angular magnitude about a 
point in several different ways. 

The method of series consists, essentially, in taking the 
readings of an angle with the circle or limb of the theodolite 
in one position, then turning the circle through an arc and 
taking the readings of the same angle again, etc. ; thus read- 
ing the angle from successive portions of the graduated circle. 

On account of the curvature of the earth, the sum of the 
three angles of a triangle upon its surface exceeds 180°. This 
spherical excess, as it is called, becomes appreciable only when 
the sides of the triangle are about 5 miles in length. To 
determine the angles of the rectilinear triangle having the 
same vertices, one-third of the spherical excess is deducted 
from each spherical angle. 



CHAPTER IV. 
LEVELLING. 



§ 30. Definitions, Curvature, and Refraction. 

A Level Surface is a surface parallel with, the surface of still 
water. 

A Level Line is a line in a level surface. 
Levelling is the process of finding the difference of level of 
two places, or the distance of one place above or below a level 
line through another place. 

The Line of Apparent Level of a place is a tangent to the level 
line at that place. ^ Hence, the line of apparent level is per- 
pendicular to the plumb-line. 

The Correction for Curvature is the deviation of the line of 
apparent level from the level line for any distance. 

Let t (Fig. 41) represent the line of apparent level of the 
t p place P, a the level line, d the diame- 

ter of the earth ; then c represents the 
correction for curvature. To compute 
the correction for curvature: 
f = c(c + d). 

Therefore, e = y—-j 

c-\-d d 

approximately, since e is very small 
compared with d, and t = a without 
appreciable error. 
Since d is constant (= 7920 miles, nearly), the correction for 
curvature varies as the square of the distance. 




THE LEVELLING ROD. 



57 



Example. What is the correction for curva- 
ture for 1 mile? 

By substituting in the formula deduced 

above, a? l!_ mi . = 8in . )near i y . 

d 7920 y 

Hence, the correction for curvature for any 
distance may be found in inches, approximate- 
ly, by multiplying 8 by the square of the dis- 
tance expressed in miles. 

A correction for the refraction of the rays of 
light is sometimes made by decreasing the cor- 
rection for curvature by one-sixth of itself. 

§ 31. The Y Level. 

This instrument is shown on page 59. 

The telescope is about 20 inches in length, 
and rests on supports called Y's, from their 
shape. The spirit level is underneath the tel- 
escope, and attached to it. The levelling- head 
and tripod are similar to the same parts of the 
transit. 

§32. The Levelling Rod. 

The rod shown in Fig. 42 is known as the 
New York levelling rod. It is made of two 
pieces of wood, sliding upon each other, and 
held together in any position by a clamp. The 
front surface of the rod is graduated to hun- 
dredths of a foot up to 6£ feet. 

The target slides along the front of the rod, 
and is held in place by two springs which press 
upon the sides of the rod. It has a square 
opening at the centre, through which the divi- 
sion line of the rod opposite to the horizontal 
line of the target may be seen. It carries a 
vernier by which heights may be read to thou- 
sandths of a foot (§ 7). 





Fig. 42. 



58 



SURVEYING. 



If a greater height than 6-J- feet is desired, the target is 
made fast with its horizontal line at 6^ feet from the lower 
end. The back part of the rod is then moved up until the 
target is at the required height, and the reading is taken at 
the side of the rod. When fully extended the rod is 12 feet 
long. 

§ 33. Difference of Level. 

To find the difference of level between two places visible from an 
intermediate place. 

Let A and B (Fig. 43) represent the two places. Set the 
Y level at a station equally distant, or nearly so, from A and 




Fig. 43. 

B, but not necessarily on the line AB. Place the legs of the 
tripod firmly in the ground, and level over each opposite pair 
of levelling screws, successively. Let the rodman hold the 
levelling rod vertically at A. Bring the telescope to bear up- 
on the rod (§ 8), and by signal direct the rodman to move the 
target until its horizontal line is in the line of apparent level 
of the telescope. Let the rodman now record the height AA' 
of the target. In like manner find BB'. The difference be- 
tween A A 1 and BB' will be the difference of level required. 
If the instrument be equally distant from A and B, or nearly 
so, no correction for curvature will be necessary. 

If the instrument be set up at one station, and the rod at 
the other, the difference between the heights of the optical 
axis of the telescope and the target, corrected for curvature 
and refraction, will be the difference of level required. 




THE Y LEVEL. 



LEVELLING FOR SECTION. 



61 



To find the difference of level of two places, one of which cannot 
be seen from the other, and both invisible from the same place; or, 
when the two places differ considerably in level, 

Let A and D (Fig. 44) represent the two places. Place the 
level midway between A and some intermediate station B. 




Fig. 44. 



Find AA' and BB\ as in the preceding case, and record the 
former as a back-sight and the latter as a, fore-sight. Select 
another intermediate station C, and in like manner find the 
back-sight BB" and the fore-sight CC ; and so continue until 
the place B is reached. The difference between the sum of the 
fore-sights and the sum of the back-sights will be the difference 
of level required. 

§ 34. Levelling for Section. 

The intersection of a vertical plane with the surface of the 
earth is called a Section or Profile. The term profile, however, 
usually designates the, Plot or representation of the section on 
paper. 

Levelling for Section is levelling to obtain the data necessary 
for making a profile or plot of any required section. 

A profile is made for the purpose of exhibiting in a single 
view the inequalities of the surface of the ground for great 
distances along the line of some proposed improvement, such 



62 SURVEYING. 



as a railroad, canal or ditch, and thus facilitating the estab- 
lishment of the proper grades. 

The data necessary for making a profile of any required 
section are, the heights of its different points above some 
assumed horizontal plane, called the Datum Plane, together 
with their horizontal distances apart or from the beginning of 
the section. 

The position of the datum plane is fixed with reference to 
some permanent object near the beginning of the section, 
called a Bench Mark, and, in order to avoid negative heights, 
is assumed at such a distance below this mark that all the 
points of the section shall be above it. 

The heights of the different points of the section above the 
datum plane are determined by means of the level and lev- 
elling-rod ; and the horizontal length of the section is meas- 
ured with an engineer's chain or tape, and divided into equal 
parts, one hundred feet in length, called Stations, marked by 
stakes numbered 0, 1, 2, 3, etc. 

Where the ground is very irregular, it may be necessary, 
besides taking sights at the regular stakes, to take occasional 
sights at points between them. If, for instance, at a point 
sixty feet in advance of stake 8 there is a sudden rise or fall 
in the surface, the height of this point would be determined 
and recorded as at stake 8.60. 

The readings of the rod are ordinarily taken to the nearest 
tenth of a foot, except on bench marks and points called 
turning points, where they are taken to thousandths of a foot. 

A Turning Point is a point on which the last sight is taken 
just before changing the position of the level, and the first 
sight from the new position of the instrument. A turning 
point may be coincident with one of the stakes, but must 
always be a hard point, so that the foot of the rod may stand 
at the same level for both readings. 
To explain the method of obtaining the field notes necessary 

for making a profile, let 0, 1, 2, 3 11 (Fig. 45) represent 

a portion of a section to be levelled and plotted. Establish a 



LEVELLING FOR SECTION. 



63 



bench, mark at or near the beginning of the line, measure the 

horizontal length of the section, and set stakes one hundred 

feet apart, numbering them 0, 1, 2, 3, . 

etc. Place the level at some point, 

as between 2 and 3, and take the 

reading of the rod on the bench 

= 4.832. Let PF represent the 

datum plane, say 15 feet below the 

bench mark, then 



15 + 4.832 - 19.832 

will be the height of the line of sight 
AP, called the Height of the Instru- 
ment, above the datum plane. . Now 
take the reading at = 5.2 = CL4, 
and subtract the same from 19.832, 
which leaves 14.6 = OP, the height 
of the point above the datum plane. <p 
Next take sights at 1, 2, 3, 3.40, * 
and 4 equal respectively to 3.7, 3.0, 
5.1, 4.8, and 8.3, and subtract the 
same from 19.832 ; the remainders 
16.1, 16.8, 14.7, 15.0, and 11.5 will 
be the respective heights of the 
points 1, 2, 3, 3.40, and 4. Then, 
as it will be necessary to change the 
position of the instrument, select a 
point in the neighborhood of 4 suit- 
able as a turning point (t.p. in the 
figure), and take a careful reading 
on it = 8.480 ; subtract this from "*» 
19.832, and the remainder, 11.352, 
will be' the height of the turning 
point. Now carry the instrument 
forward to a new position, as between 5 
figure, while the rodman remains at t.p. ; 



and 6, shown in the 
take a second reading 



64 



SURVEYING. 



on t.p. — 4.102, and add it to 11.352, the height of t.p. above 
PP 1 ; the sum 15.454 will be the height of the instrument 
CD in its new position. Now take sight upon 5, 6, and 7, 
equal respectively to 4.9, 2.8, and 0.904 ; subtract these sights 
from 15.454, and the results 10.6, 12.7, and 14.550 will be the 
heights of the points 5, 6, and 7 respectively. The point 7, 
being suitable, is made a turning point, and the instrument is 
moved forward to a point between 9 and 10. The sight at 
7 = 6.870 added to the height of 7 gives 21.420 as the height 
of the instrument EF in its new position. The readings at 
8, 9, 10, and 11, which are respectively 5.4, 3.6, 5.8, and 9.0, 
subtracted from 21.420, will give the heights of these points, 
namely, 16.0, 17.8, 15.6, and 12.4. 

Proceed in like manner until the entire section is levelled, 
establishing bench marks at intervals along the line to serve 
as reference points for future operations. A record of the work 
described above is kept in a field book as follows : 



Station. 


+ S. 


H.I. 


-s. 


H.S. 


Remarks. 


B 


4.832 






15. 


Bench on rock 20 ft. 







19.832 


5.2 


14.6 


sonth of 0. 


1 






3.7 


16.1 




2 






3.0 


16.8 




3 






5.1 


14.7 


3 to 3.40 turnpike road. 


3.40 






4.8 


15.0 




4 






■8.3 


11.5 




t.p. 


4.102 




8.480 


11.352 




5 




15.454 


4.9 


10.6 




6 






28. 


12.7 


_ 


7 


6.870 




0.904 


14.550 




8 




21.420 


5.4 


16.0 




9 






3.6 


17.8 




10 






5.8 


15.6 




11 






9.0 


12.4 




B 










Bench on oak stump 


12 










27 ft. N.E. of 12, 


etc. 










etc. 



LEVELLING FOR SECTION. 65 

The first column contains the numbers or names of all the 
points on which sights are taken. The second column con- 
tains the sight taken on the first bench mark, and the sight on 
each turning point taken immediately after the instrument 
has been moved to a new position. These are called Plus 
Sights (+/&) because they are added to the heights of the 
points on which they are taken to obtain the height of the 
instrument given in the third column (H.I.). The fourth 
column contains all the readings except those recorded in the 
second column. These are called Minus Sights (— 8.) because 
they are subtracted from the numbers in the third column to 
obtain all the numbers in the fifth column except the first, 
which is the assumed depth of the datum plane below the 
bench. The fifth column contains the required heights of all 
the points of the section named in the first column together 
with the heights of all benches and turning points. 

To find the difference of level between any two points of 
the section, we have only to take the difference between the 
numbers in the fifth column opposite these points. 

The real field notes are contained in the first, second, fourth, 
and last columns ; the other columns may be filled after the 
field operations are completed. The field book may contain 
other columns; one for height of grade (H.G.), another for 
depth of cut (C), and another for height of embankment or fill 
(F.). 

To plot the section. Draw a line PP (Fig. 45), to repre- 
sent the datum plane, and beginning at some point as P, lay 
off the distances 100, 200, 300, 340, 400 feet, etc., to the right, 
using some convenient scale, say 250 feet to the inch, as in the 
figure. At these points of division erect perpendiculars equal 
in length to the height of the points 0, 1, 2, 3.40, 4, etc., 
given in the fifth column of the above field notes, using in 
this case a larger scale, say 20 feet to the inch. Through the 
extremities of these perpendiculars draw the irregular line 

0, 1, 2, 3 11, and the result, with some explanatory figures, 

will be the required plot or profile. 



66 



SURVEYING. 



The making of a profile is much simplified by the use of 
profile paper, which may be had by the yard or roll. 

If a horizontal plot is required, the bearings of the differ- 
ent portions of the section must be taken. 



§ 35. Substitutes for the Y Level. 

For many purposes not requiring accuracy, the following 
simple instruments in connection with a graduated rod will be 
found sufficient. 



&==§- 



Fig. 46. 



Fig. 47. 



Fig. 48. 



The Plumb Level (Fig. 46) consists of two pieces of wood 
joined at right angles. A straight line is drawn on the up- 
right perpendicular to the upper edge of the cross-head. 

The instrument is fastened to a support by a screw through 
the centre of the cross-head. The upper edge of the cross-head 
is brought to a level by making .the line on the upright coin- 
cide with a plumb-line. 

A modified form is shown in Fig. 47. A carpenter's square 
is supported by a post, the top of which is split or sawed so as 
to receive the longer arm. The shorter arm is made vertical 
by a plumb-line which brings the longer arm to a level. 

The "Water Level is shown in Fig. 48. The upright tubes are 
of glass, cemented into a connecting tube of any suitable ma- 
terial. The whole is nearly filled with water, and supported 
at a convenient height. The surface of the water in the up- 
rights determines the level. 

By sighting along the upper surface of the block in which 
the Spirit Level is mounted for the use of mechanics, a level 
line may be obtained. 



SUBSTITUTES FOE, THE Y LEVEL. 



67 



Exercise V. 

1. Find the difference of level of two places from the fol- 
lowing field notes : back-sights, 5.2, 6.8, and 4.0 ; fore-sights, 
8.1, 9.5, and 7.9. 

2. Write the proper numbers in the third and fifth columns 
of the following table of field notes, and make a profile of the 
section. 



Station. 


+ s. 


H.I. 


-S. 


H.S. 


Remarks. 


B 


6.944 






20. 


Bench on- post 22 ft. 









7.4 




north of 0. 


1 






5.6 






2 






3.9 






3 






4.6 






t.p. 


3.855 




5.513 






4 






4.9 






5 






3.5 






6 






1.2 







3. Stake of the following notes stands at the lowest point 
of a pond to be drained into a creek ; stake 10 stands at the 
edge of the bank, and 10.25 at the bottom of the creek. Make 
a profile, draw the grade line through and 10.25, and fill 
out the columns H.G. and C, the former to show the height of 
grade line above the datum, and the latter, the depths of cut 
at the several stakes necessary to construct the drain. 



Station. 


+ s. 


H.I. 


-s. 


H.S. 


H.G. 


c. 


Remarks, 


B 


6.000 






25. 






Bench on rock 









10.2 




20.8 


0.0 


30 feet west of 


1 






5.3 






5.3 


stake 1. 


2 






4.6 










3 






4.0 










4 






6.8 










5 


4.572 




7.090 










6 






3.9 










7 






2.0 










8 






4.9 










9 






4.3 










10. 






4.5 










10.25 






11.8 











68 



SURVEYING. 



§ 36. Topographical Levelling. 

The principal object of topographical surveying is to show 
the contour of the ground. This operation, called topographi- 
cal levelling, is performed by representing on paper the 
curved lines in which parallel horizontal planes at uniform 
distances apart would meet the surface. 

It is evident that all points in the intersection of a horizontal 
plane with the surface of the ground are at the same level. 
Hence, it is only necessary to find points at the same level, 
and join these to determine a line of intersection. 

The method commonly employed will be understood by a 
reference to Fig. 49. The ground ABCD is divided into 
equal squares, and a numbered stake driven at each intersec- 
tion. By means of a level and levelling rod the heights of the 
other stations above m and J), the lowest stations, are deter- 
mined. A plot of the ground with the intersecting lines is 

then drawn, and the height of 
each station written as in the 
figure. 

Suppose that the horizontal 
planes are 2 feet apart ; if the 
first passes through ra and D, 
the second will pass through p, 
which is 2 feet above ra; and 
since n is 3 feet above ra, the 
second plane will cut the line 
ran in a point s determined by 
the proportion ran : ras : : 3 : 2. 
In like manner the points t, q, 
and r are determined. 

The irregular line tsp qr represents the intersection of 

the second horizontal plane with the surface of the ground. 
In like manner the intersections of the planes, respectively, 
4, 6, and 8 feet above ra, are traced. 

It is evident that the more rapid the change in level the 
nearer these lines will approach each other. 




Fig 49. 



CHAPTER V. 
RAILROAD SURVEYING. 



§ 37. General Remarks. 

When the general route of a railroad has been determined, 
a middle surface line is run with the transit. A profile of this 
line is determined, as in § 34. The levelling stations are com- 
monly 1 chain (100 feet) apart. Places of different level are 
connected by. a gradient line, which intersects the perpendic- 
ulars to the datum line at the levelling stations in points 
determined by simple proportion. Hence, the distance of each 
levelling station, above or below the level or gradient line 
which represents the position of the road bed, is known. 



§ 38. Cross Section Work. 




Excavations. If the road bed lies below the surface, an exca- 
vation is made. 

Let ACDB (Fig. 50) represent a cross section of an excava- 
tion, /a point in the middle surface line, /' the corresponding 
point in the road bed, and CD the width of the excavation at 
the bottom. The slopes at the sides are commonly made so 



70 



SURVEYING. 



that AA' = iA'C, and BB' = %DB'. //' and CD being 
known, the points A, B, C\ and D' are readily determined by 
a level and tape measure. 

If from the area of the trapezoid ABB 1 A 1 the areas of the 
triangles J.J.'(7and BB'B be deducted, the remainder will be 
the area of the cross section. 

In like manner the cross section at the next station may be 
determined. These two cross sections will be the bases of a 
frustum of a quadrangular pyramid whose volume will be the 
amount of the excavation, approximately. 

Embankments. If the road bed lies above the surface, an 
embankment is made, the cross section of which is like that of 
the excavation, but inverted. 




Fig. 51 represents the cross section of an embankment 
which is lettered so as to show its relation to Fig. 50. 



§ 39. Kailroad Curves. 

When it is necessary to change the direction of a railroad, 
it is done gradually by a 
curve, usually the arc of 
a circle. 

Let AB and AO (Fig. 
52) represent two lines to 
be thus connected. Take 
AD = AE=t. The in- 
tersection of the perpen- 
diculars DC and EC de- 
termines the centre C, and 
the radius of curvature DC=r. The length of the radius 




RAILROAD CURVES. 



71 



of curvature will depend on the angle A and the tangent AD. 
For, in the right triangle ADC, 



tan 2X4 C= 
Hence, r- 



— — -, or tan + ^1 = — 
AD' 2 t 

t tan -J J.. 



First Method, 




To Lay out the Curve. 

Let Dm (Fig. 53) represent a portion of the 
tangent. It is required to find mP, the 
perpendicular to the tangent meeting the 
curve at P. 

mP=Dn = CD-Cn. 

CD = r, _ 
Cn = ^CP' i 
= Vr^~ 



Jff 



■f. 



Fig. 53. 



Hence, 



mP^r-Vr 2 -* 2 . 




Second Method. It is required to find 
mP (Fig. 54) in the direction of the 
centre. 

mP = mC-PC. 



But mC = ^DC 2 + Dm 2 = Vr 2 -f t 2 . 
Hence, 



Fig. 54. 

Third Method. 



m 



P = -y/i*+f 



r. 




Fig. 55 



Place transits at D and E (Fig. 55). Direct 
the telescope of the former 
to 27, and of the latter to A. 
Turn each toward the curve 
the same number of degrees, 
and mark P, the point of 
intersection of the lines of 
sight. P will be a point in 



the circle to which AD and 
AE are tangents at D and E, respectively. 



72 



SURVEYING. 



Fourth Method. Measure the angle A and the tangents 
AD = A£(Fig.b6). Com- 



pute r. Select any distance 
I (the length of the chain 
will be found convenient). 
Place the transit at D and 
direct toward A. Turn off 
an angle ADP determined 
by the equation sin ADP 




Fig. 56. 



I 



and measure DP=l\ make the angle ADP' = 2 ADP 
make the angle ADP" = S ADP and P'P" 



2r 

and PP = l; 

= I; etc. P } P 1 , P", etc., will be in the circumference of the 

required arc. 



ANSWEES. 



PLANE TRIGONOMETRY. 

Exercise I. 

1. sinJS = -, cosB = -, tan£ = _, cotB = ^ t secB = ~, cscB = 
c c aba 

3. (i.) sin = f , cos = -f, (ii.) sin = T 5 ¥ , etc. (v.) sin = f f , etc. 
tan = f, cot = |, (iii.) sin = T 8 T , etc. (vi.) sin = |£§ , etc. 



m 



4. The required condition is that a 2 + b 2 = c 2 . It is. 

5. (i.) sin = -v — 5, etc. (in.) sin = -, etc. 
v ' m 2 + n 2 s 

(ii.) sin = , xy , , etc. (iv.) sin = — , etc. 

v ' x 2 + y 2 qr 

7. In (iii.) p 2 q 2 + q 2 s 2 = p 2 s 2 ; in (iv.) ra 2 n 2 s 2 + m 2 p 2 v 2 = n 2 q 2 r 2 . 

8. c= 145 ; whence, sin A = T 2 ^ = cos B ; cos A = ££f = sin B ; 
tan J. = T 2 ¥ 4 j = cot B ; cot J. = *££ = tan 5 ; sec A = iff = esc i? ; etc. 

9. b = 0.023 ; whence, tan A = cot 5 = ,& 3 ¥ ; cot^i = tan B = ^*, etc. 

10. a = 16.8 ; whence, sin A = jff = cos B, etc. 

11. c = » + <7; whence, sin A = —+■ — = cosi?; etc. 



12. b = Vq{p + q) ; whence, tan A = -vl- = cot 5 ; etc. 

13. a = » — a; whence, sin A = — ; — = cosB: etc. 

14. sinJ. = fV5 = 0.89443; etc. 15. sin^i = § ; etc. 

16. sin A = $ (5 + V7) - 0.95572 ; etc. 

17. cost! = £(v^1-1) = 0.57097; sin 4 = &(V31 + 1) = 0.82097; etc. 

18. a =12.3. 20. a = 9. 22. c=40. 

19. 6 = 1.54. 21. 6 = 68. 23. c = 229.62. 

24. Construct a rt. A with legs equal to 3 and 2 respectively; then 
construct a similar A with hypotenuse equal to 6. 
In like manner, 25, 26, 27, may be solved. 
28. a = 1.5 miles; 6 = 2 miles. 31. 400,000 miles. 

30. a = 0.342, 6 = 0.940; a =1.368, 6 = 3.760. 32. 142.926 yards. 



TRIGONOMETRY. 



Exercise II. 

5. Through A (Fig. 3) draw a tangent, and take A T= 3 ; the angle 

AOT is the required angle. 

6. From (Fig. 3) as a centre, with a radius = 2, describe an arc cut- 

ting at 8 the tangent drawn through B ; the angle SO A is the 
required angle. 

7. In Fig. 3, take GW= £, and erect MP ± OA and intersecting the 

circumference at P; the angle POM is the required angle. 

8. Since sin x = cosz, OM= PM (Fig. 3), and x = 45° ; hence, construct 

z = 45°. 

9. Construct a rt. A with one leg = twice the other ; the angle opposite 

the longer leg is the required angle. 

10. Divide OA (Fig. 3) into four equal parts ; at the first point of divi- 
sion from erect a perpendicular to meet the circumference at 
some point P. Join OP; the angle AOP is the required angle. 

21. r sin re. 22. Leg adjacent to A = nc, leg opposite to A = mc. 



Exercise III. 

1. cos 60°. cotl°. sec 71° 50'. tan 7° 41'. 
sin 45°. tan 75°. sin 52° 36'. sec 35° 14'. 

2. cos 30°. cot 33°. sec 20° 58'. tanO 1'. 
sin 15°. tan 6°. sin 4° 21'. sec 44° 59. 

3. £V3. 

4. tan A = cot A = cot (90° - A) ; hence, A = 90° - A and A = 45°. 

5. 30°. 7. 90°. 9. 22° 30'. 11. 10°. 

6. 30° 8. 60°. 10. 18°. 12 90° 

n + 1 

Exercise V. 

1. coaA = T 5 - s , tan 4 = -^, cotJ. = T 5 ^, secA=^-, cscA={%. 

2. cos A = 0.6, ha®A = 1.3333, cot^ = 0.75, sec^ = 1.6667, esc J.=1.25. 

3. sin -d. = -H, tanJ. = £i cofcJ. = ff sec J. = f £, cscJ.= ff 

4. sin A = 0.96, tan A = 3.42854, cot J. = 0.29167, sec A = 3.5714. 

5. sin A = 0.8, cos A = 0.6, cot A = 0.75, sec J. = 1.6667, esc 4 = 1.25. 

6. sin A = %V2 } cos A=%V2 f tanJ.= l, secA=V2, cscA = V2. 

7. tan A =- 2, sin 4 = 0.90, cos 4 = 0.45, sec 4 = 2.22, esc A = 1.11. 



ANSWERS. 



cos 



,A = i, smA=$y/3, tan^i=V3, cotJ.=£V3, csc^i = fV3. 
9. sin A = lV2, cos J. = J\/2, tan A = l, cot A = 1, sec J. = V2. 

10. cos^=Vl-m 2 , tanvi = -^-Vl-ra 2 , cot^-V^Z^ 

1 — wr m 

11. cosX- 1 -^. tanA = -J»> coti.= i^, MC.l-i+4 

1 + m 2 1 — m 2 2m 1 — m 3 

12. sin A = > tan A = — > sec .a = — • 

m 2 + n 2 2 raw 2 mn 

13. cot = 1, sin = £ V2, cos = %V2, sec =V2, esc =V2. 

14. cos = £V3, tan = ^V3", cot=V3, sec = |V3, esc = 2. 

15. sin = £V3, cos=$, tan=V3, cot = |V3, sec = 2. 

16. sin-^V2-v / 3, cos = jV2 +V3, cot = 2+V£ 

17. sin = ^V2-V2, cos = ^V2+V2, tan = V2-l. 

18. cos = 1, tan = 0, cot = oo, sec = 1, esc = oo. 

19. cos = 0, tan = oo, cot = 0, sec = oo, esc = 1. 

20. sin = 1, cos = 0, cot = 0, sec = oo, esc = 1. 

21. cosA=Vl-sm 2 A, tan^l = — g^ ■ csc^L = — L_. 

Vl-sinM sin J. 

22. sin^iWl-cosM, tan^=^!H™ZI, cot ^ = cos A 

cosJ - Vl-cosM 

sec A — 1 esc A 



cos A Vl-cosM 

23. sin A = — jjg^j cos^ = 1 ca/tA. \ . 

Vl+tanM VI + tan 2 J. tan -4 

sec A = Vl+tanM, esc A = ^ /l + tap2 A 

tan A 

24. tan A = -L-, esc A = Vl + cot 2 A, sin A = l 

cotA Vl+cot 2 ^ 

cos,l= ™ iA , secA= VTT^A t 
VI + cot 2 A cot A 

25. BmA = ±VE, cos^=|V5. 27. smA = ^ T , cos^ = ff 

26. sin A = 1^/15, tan.A=Vl5. 2g 1 - 3cosM + 3 cos* A 

con 2 A — cos 4 A 



TRIGONOMETRY. 





Exercise VI. 


1. 


6 . 6 5. 
- = cos A\ r.e = . 

c : : : A 


A = 90° -B, a = ccosB, 
6 = c sin B. 




39. c= 7.8112, .4 = 39° 48', 


B = 5Q° 12'. 




40. 6 = 70.007, 4 = 30* 12" 


£= 89° 29' 48". 




41. n= 1.1886, A = 43° 20', 


B = 46° 40'. 




42. £ = 21.249, c = 22.372, 


£=71° 46'. 




43. a =6.6882, c = 13.735, 


B = 60° 52'. 




44. a =63.89, 5 = 23.369, 


5 = 20° 6'. 




45. a =19.40, 6 = 18.778, 


JL = 45° 56'. 




46. 6 = 53.719. c= 71.377. 


JL = 41° 11'. 




47. a =12.981, c = 15.796, 


JL = 55° 16'. 




48. a=0.5S046, 6 = 8.442, 


A= 3° 56'. 


49. 


i 7 = | (c 2 sin J. cos J.). 5 1 . 


F = 1(6* tan A). 


50. 


F=$(a* cot A). 52. 


F==|(oVc»-o«). 


53. 


6 = 11.6, c= 15.315, J. = 40° 45' 48", B = 49° 14' 12". 


54. 


a = 7.2, c = 8.766, £ = 34° 46' 40", 4 = 55° 13' 20". 


55. 


a = 3.6474, 6 = 6.58, c = 7.5233. 


£ = 61°. 


56. 


a = 10.283, 6 = 19.449, -4 = 27=52' 


B = 62° 8'. 


57. 


19° 28' and 70° 32'. 65. 


tan JL = ^, .4 = 59° 45'. 
6 

a = 6tanJ., 95.34. 


58. 


3 and 5.1961. 

66. 

cr.~ 
a = ccos r 67. 

n - 1 

90° 68. 

b = c sm 

n + 1 69. 


59. 


1 = 25' 56". 

7.0712 miles in each direction 
20.88 feet 


60. 


36= 52' 12" and 53° 7' 48". 70. 


56.65 feet 


61. 


212.1 feet. 71. 


228.63 yards. 


62. 


732.22 feet 72. 


136.6 feet 


63. 


-"feet. 73. 


140 feet. 


64. 


37.3 feet, 96 feet. 74. 


84.74 feet. 



Exercise VII. 

1. C = 2( 90°-JL), c = 2acosrA. h = asmA. 

2. A=%(WP-(J) l c = 2acosJL, fi = asinJL 

3. C = 2(90°-JL), a = c + 2cosA,h = a^A. 





ANSWERS. 5 


4. 


J.= £(180°-C), a = c -4- 2 cos J., A = asinJ.. 


5. 


C«=2 (90° — A), a = h+ sin A, c = 2a cos A 


6. 


J. = J (180° - C\ a = h -h sin A, c = 2a cos J.. 


7. 


sin A = h + a, 0= 2(90° - A), c = 2 a cos A. 


8. 


tan4=A-s-£c, C=2(90 o -u4), <z = A -=- sin A 


9. 


J. = 67° 22' 50", C= 45° 14' 20", h = 13.2. 


10. 


c = 0.21943, h = 0.27384, i?= 0.03004. 


11. 


a = 2.055, A = 1.6852, F= 1.9819. 


12. 


a = 7.705, c = 3.6676, F= 13.73. 


13. 


J. = 79° 36' 30", 0= 20° 45' 30", c = 2.4206. 


14. 


A = 77° 19' 11", C= 25° 21' 38", a = 20.5. 


15. 


A = 25° 28', C=129°4', a = 81.388, A = 35. 


16. 


A = 81° 12' 9", (7= 17° 35' 42", a = 17, c = 5.2. 


17. 


F=\cy/^a?~c\ 22. 0.76537. 


18. 


F=a?smiCco8%C. 23. 94° 20'. 


19. 


.F= a 2 sin .A cos A 24. 2.7261. 


20. 


F=hn&n%a 25. 38° 56' 33". 


21. 


28.284 feet, 4525.44 sq. feet. 26. 37.7 



Exercise YIII. 



1. 


r = 1.618, h = 1.5388, J P= 7.694. 






2. 


r = 11.269, A = 10.886, .F= 381.04. 






3. 


h = 0.9848, p = 6.2514, F= 3.0782. 






4. 


A = 19.754, c = 6.2537, i^= 1236. 






5. 


r = 1.0824, c = 0.8284, jP= 3.3137. 






6. 


r = 2.592, h = 2.488, c= 1.4615. 






7. 


r= 1.5994, ft = 1.441, p = 9.716. 






8. 


0.6181. 12. 0.2238. 


17. 


11.636. 


9. 


0.64984. 13. 0.31. 


18. 


99.64. 


10. 


0.51764. 14. 0.82842. 


19. 


1.0235. 


11. 


h c 15. 94.63. 


20. 


0.635. 




2,ns^ 16. 415. 





TRIGONOMETRY. 



Exercise IX. 

5. Two angles: one in Quadrant I., the other in Quadrant II. 

6. Four values : two in Quadrant I., two in Quadrant IV. 

7. x may have two values in the first case, and one value in each of 
the other cases. 

8. If cosa; = — f, x is between 90° and 270° ; if cot a; = 4, x is between 
0° and 90° or 180° and 270° ; if sec x = 80, x is between 0° and 
90° or between 270° and 360°; if cscz = -3, x is between 180' 
and 360°. 

9. In Quadrant III. ; in Quadrant II. ; in Quadrant III. 

10. 40 angles ; 20 positive and 20 negative. 

11. +, when x is known to be in Quadrant I. or IV. ; — , when x is known 
to be in Quadrant II. or III. 

14. sin# = — fV3, tancc = — 4V3, cot x = — ^V3, esc a; = — -j^VS. 

15. sino; = + y^VlO, cosa: = + T 3 ^VI0, tan x = — J, sec x = + £ VTO. 

cscrr = ± VlO. 

16. The cosine, the tangent, the cotangent, and the secant are negative 
when the angle is obtuse. 

17. Sine and cosecant leave it doubtful whether the angle is an acute 
angle or an obtuse angle ; the other functions, if + determine an 
acute angle, if — an obtuse angle. 

20. sin 450 = sin (360 + 90) = sin 90 = 1 ; tan 540° = tan 180° = ; 
cos 630° = cos 270° = ; cot 720° = cot 0° = 00 ; 
sin 810° = sin 90° = 1 ; esc 900° = esc 180° = 00. 

21. 45°, 135°, 225°, 315°. 22. 0. 23. 0. 24. 0. 

25. a 2 -Z> 2 + 4aZ>. 

Exercise X. 

2. sin 172°- sin 8°. 11. cot 264° = tan 6°. 

3. cos 100° = - sin 10°. 12. sec 244° = - esc 26°. 

4. tan 125° --cot 35°. 13. esc 271° = - sec 1°. 

5. cot 91° --tan 1°. 14. sin 163° 49' = sin 16° 11'. 

6. sec 110° =- esc 20°. 15. cos 195° 33' = - cos 15° 33'. 

7. esc 157°= esc 23°. 16. tan 269° 15'= cot 0° 45'. 

8. sin 204° = - sin 24°. 17. cot 139° 17' = - cot 40° 43'. 

9. cos 359°- cos 1°. 18. sec 299° 45'= esc 29° 45'. 
10. tan 300°= -cot 30°. 19. esc 92° 25' = sec 2° 25'. 



ANSWERS. 



20. sin(- 75°)= -sin 75°= -cos 15°, cos(- 75°)= cos 75°= sin 15°, etc. 

21. sin(-127°)= -sin 127°= -cos 37°, cos(-127°)= cos 127°= -sin 37°, etc. 

22. sin(-200°)= sin 160°= sin20 o ,cos(-200 o )=cos200 o =-cos20° > etc. 

23. sin( -345°) = -sin 345° = sin 15°, cos(-345°) = cos 345° = cos 15°. etc. 

24. sin(- 52° 37') = -sin 52° 37' = - cos 37° 23', 
cos (- 52° 37') = cos 52° 37' = sin 37° 23', etc. 

25. sin (- 196° 54') = sin 196° 54'= cos 16° 54', 
cos (- 196° 54') = cos 196° 54' = - cos 16° 54', etc. 

26. sin 120° = § V3, cos 120° = -£, etc. 

27. sin 135° = + £ V2, cos 1 35° = - J V2, etc. 

28. sin 150° = + l cos 150° = - £ V3, etc. 

29. sin 210° = - J, cos 210° = - § V3, etc. 

30. sin 225° = - | V2, cos 225° = - \ V2, etc. 

31. sin 240° = - £\/3, cos 240° = - |, etc. 

32. sin 300° = -|Vf, cos 300° = + J, etc. 

33. sin (-30°) = -£, cos (-30°) = + \ V3, etc 

34. sin (- 225°) = + %V2, cos(- 225°) = - \ V2, etc. 

35. cos x = - J V2 or - vj, etc., a; = 225°. 

36. tan a; = — V|~ sin a? = |, cos x = — J \/3, a; = 150°. 

37. sin 3540° = sin 300° = - sin 60° = - \ V3, cos 3540° *- + |, etc. 

38. 210° and 330° ; 120° and 300°. 

39. 135°, 225°, and -225° ; 150° and -30°. 

40. 30°, 150°, 390°, and 510°. 

41. sin 168°, cos 334°, tan 225°, cot 252°, 
sin 349°, cos 240°, tan 64°, cot 177°. 

42. 0.848. (Hint: tan 238° = tan 58°, sin 122° = sin 58°). 

43. -1.952. 47. a 2 + Z> 2 + 2 ab cos x. 

44. (a — b) sin x. 48. 0. 

45. m sin x cos x. 49. cos x sin y — sin a; cos y. 

46. (a — b) cot a? - (a + b) tan a. 50. tan x. 

51. Positive between a? = 0° and a;=135°, and between x = 315° and 

a? = 360° ; negative between x = 135° and x = 315°. 

52. Positive between x = 45° and x = 225° ; negative between x = 0° 

and a: = 45°, and between x = 225° and x = 360°. 

53. sin (x — 90°) = — cos x, cos (x — 90°) = sin x, etc. 

54. sin (x — 180°) = — sin x, cos (a? — 180°) = — cos x, etc. 

Exercises 53 and 54 should be solved by drawing suitable figures, and 
employing a mode of proof similar to that used in \ 28. 



8 



TRIGONOMETRY. 



Exercise XI. 

1. sin (x + y) = ff, cos (a; + y) = ff. 

3. sin (90° + y) = cosy, cos( 90 +y) = — siny, etc. 

4. sin (180— y) = sin y, cos (180 —y) = —cosy, etc. 

5. sin (180 + y) = —sin y, cos (180 + y) = —cos y, etc. 

6. sin (270 — y) = — cos y, cos (270 — y) = — sin y, etc. 

7. sin (270 + y) = — cos y, cos (270° + y)= sin y , etc. 

8. sin (360° - y)= - sin y , cos (360 - y) = cos y, etc. 

9. sin (360 +y)= sin y, cos (360° + y)= cos y, etc. 

10. sin (x - 90°) = - cos x, cos (a? - 90°) = sin x, etc. 

11. sin (x — 180°)= — sin x, cos (x — 180°)= — cos re, etc. 

12. sin (x -270°)= cosx, cos (x- 270°)= -sin x, etc. 

13. sin(— y) =— siny, cos(— y) = cosy, etc. 

14. sin(45°— y) = £ V2(cosy— siny), cos(45°— y) = J v / 2(cosy+siny),etc. 

15. sin(45°+y) = £\/2(cosy+siny), cos(45°+y) = J V2 (cosy— siny), etc. 

16. sin (30°+y) = £ (cos y+V3 siny), cos(30 o +y) = £(V3 cosy— siny), etc. 

17. sin (60°— y) — J(V3~cosy— siny), cos (60°— y) = £ (cos y+V3 siny), etc. 

18. 3 sin x- 4 sin 3 a. 19. 4 cos 3 re- 3 cos x. 20. 0. 21. $V3. 



22. sin £re =^| 1 ~ 0A ^ 6 = 0.10051 ; cosfrs =<J 1 + 0Ay/& = 0.99494. 

23. cos2re = -£, tan2re = — V3. 

24. sin 22£° = iV2-V2 = 3827, cos 22£° = jV2+V2 = 0.9239. 
tan 22 £° = V2 -1 = 0.4142, cot 22£° = V2 + 1 = 2.4142. 

25. sin 15° = J V2-V3 = 0.2588, cos 15° = J V2 +V3 = 0.9659. - 
tan 15° = 2 - \/3 = 0.2679, cot 15° = 2 + V3 = 3.7321. 

27-33. The truth of these equations is to be established by expressing 
the given functions in terms of the same function of the same 
angle. Thus, in Example 27, 
sin 2x = 2 sin a; cos re, 

and 2tanrc=2 , 1 + tan 2 rc = sec 2 rc = — =-*-• 

cos re cos 2 re 

By making these substitutions in the given equation its truth 
will be evident. 
34. sin A + sin B + sin C= sin A + sin B + sin [180 ~(A + B)] 
— sin A + sin B + sin (A + B) 

= 2 sin £ (A + B ) cos|(A - B) + 2sin£(A + B)cos%(A + B) 
= 2 sin J (A + B) [cos£(A - B) + cos£(A + B)} 
= 4sini(A + B) cos J A cos %B, (see gg 34 and 35) 

But cos £ C = cos [90° - £ ( A + £)] = sin | ( A + B). 

Therefore, sin A + sin B + sin C= 4 cos \ A cos £ B cos £ C. 



ANSWERS. 



35. Proof similar to that for 34. 

nn . . jy , , n sin A cos B cos A sin B sin 

36. tan A + tan B + tan C = — — - + - + 

cos A cos B cos A cos B cosC 

sin C sin C sin C cos 0+ cos A cos i? sin C 



+ 



cos J. cos B cosC cos J. cos B cos (7 

(cos AcosB + cos C) sin C_ [cos AcosB — cos (A. + B)] sin C 
cos J. cos B cos (7 cos A cos i? cos C 

z? «,-^ /^ 

= tan A tan 5 tan C. 



sin A sin i? sin 



cos A cos I? cos C 

37. Proof similar to that for 36. 

o q 2 42. tan 2 a. .„ cos (a; + y) 

sin 2a? .~ coa(x—y) sin a; sin y 

39. 2 cot 2a;. ' cos a; cosy 47. tan a; tan y. 

4a cos(a;-y) 44 cosfo+y) 

sin x cos y cos x cos y 

4L cos(a-f y) 45< cos (a; -y) 

sin a; cos y sin a? sin y 

Exercise XII. 

1. If, for instance, B= 90°, [25] becomes J = sin A. 

o 
3. a 2 = b 2 + c 2 , a 2 = b 2 + c 2 -2bc, a 2 = b 2 + c 2 + 2bc. 

6. 90° in each case. 

7. (i.) a ~ = tan (J. — 45°), and a right triangle. 

a + b 

(ii.) a + b = (a — b) (2 +V3), an isosceles triangle with the angles 30 c 
30°, 120°. 

Exercise XIII. 

9. 300. 15. a = 5, e = 9.6592. 

10. AB = 59.564 miles ; 16. a = 7, 6=8.573. 

AC = 54.285 miles. 1? gideg) 60Q feet and 1Q39 2 fget , 

11. 4.6064 miles, 4.4494 miles, altitude, 519.6 feet. 
3.7733 miles. 



12. 4.1501 and 8.67. 

13. 6.1433 miles and 8.7918 miles. 

14. 8 and 5.4723. 



18. 855:1607. 

19. 5.438 and 6.857. 

20. 15.588. 



Exercise XIV. 

11. 420. 12. The other diagonal = 124.617 



10 






TRIGONOMETRY. 








Exercise XV. 


11. 


6. 


15. 


25. 18. 10.266. 


12. 


10.392. 


16. 


3800 yards. 19. a = 5.0032, b = 2.3385. 


14. 


8.9212. 


17. 


729.7 yards. 20. 26° 0' 10" and 14° 5' 50". 
Exercise XVI. 



11. A = 36° 52f 12", B = 53° 7' 48", 0= 90°. 16. 45°, 60°, 75°. 

12. A = B = 33° 33' 27", 0= 112° 53' 6". 17. 4° 23' W. of N., or W. of S. 

13. ^i = £=<?= 60°. 18. 60°. 

14. Impossible. 20. 0.88873. 

15. 15°, 45°, 120°. 21. 54.516 miles. 



Exercise XVII. 



1. 4333600. 

2. 365.68. 

3. 13260. 

4. 8160. 

5. 240. 

6. 26208. 

7. 15540. 

8. 29450 or 6983. 



9. 10V3 = 17.3205. 

10. 6V3 = 10.3923. 

11. 0.19952. 

12. ab sin A. 

13. i(a 2 -J 2 )tanX 

14. 2421000. 

15. 30°, 30°, 120°. 



1. 21.166 miles 

2. 6.3399 miles. 

3. 119.29 feet. 

4. 30°. 



Exercise XVIII. 
24.966 miles. 



5. 20 feet. 

6. 2.6268 or 21.4704, 

7. 276.14 yards. 

8. 383.35 yards. 



ANSWERS. 11 



Miscellaneous Problems. 



2. 


107 feet; 143 feet. 27. 


8 inches. 


51. 


757.5 feet. 


3. 


1024 feet. 


30. 


460.45 feet. 


52. 


520 yards. 


4. 


37° 44' 5". 


31. 


88.94 feet. 


53. 


1366.4 feet. 


5. 


238,400 miles. 


32. 


13.657 miles. 


54. 


658 pounds ; 


6. 


861,800 miles. 


34. 


56.5 feet. 




22° 24' with first 


7. 


2922.4 miles. 


35. 


51.6 feet. 




force. 


8. 


60°. 


36. 


101.89 feet. 


55. 


88.33 pounds ; 


9. 


3.2. 


38. 


N. 76° 56' E. ; 




45° 37' with known 


10. 


6.6. 




13.94 miles an 


hour. 


force. 


11. 


199.56 feet. 


39. 


422 yards. 


58. 


500.2; 536.3. 


12. 


43 feet. 


40. 


256 feet. 


59. 


345.47 feet. 


13. 


45 feet. 


41. 


3121 feet ; 


60. 


345.47 yards. 


14. 


26° 34'. 




3633.5 feet. 


61. 


61.23 feet. 


15. 


78.37 feet. 


42. 


529.5 feet. 


63. 


307.79. 


16. 


75 feet. 


43. 


41.41 feet. 


64. 


19.8; 35.7; 44.5. 


17. 


1.44 miles. 


44. 


234.5 feet. 


65. 


45°. 


19. 


56.65 feet. 


45. 


25.43 miles. 


68. 


60°. 


20. 


69.28 feet. 


46. 


294.7 feet. 


69. 


60°. 


21. 


260 feet ; 3690 feet. 47. 


12,492 feet. 


70. 


30°. 


22. 


1.344 miles. 


48. 


6.34 miles. 


73. 


£ be sin A. 


23. 


235.8 yards. 


49. 


210.44 feet. 








74. 


\ c 2 sin A sin B esc (.4 


+ B). 






75. 


vl«0 


> — a) (s — b) (s — 


o)l 




77. 


199 A. 3 e. 10 p. 


94. 


16,281. 


114. 


S. 56° 7' 30" E. : 


78. 


210 a. 3 r. 26 p. 


95. 


435.8 feet. 




202.6 miles. 


79. 


12 a. 3 r. 37 p. 


96. 


49,089 feet. 


115. 


N. 17°25'W.; 


80. 


3 a. r. 6 p. 


97. 


750 feet. 




37° 46' N. 


81. 


12 A. 1 R. 14 P. 


98. 


422.4 feet. 


116. 


56°11'E. ; 244.3 


82. 


4 a. 2 b. 26 p. 


99. 


1835 feet. 


121. 


Long. 68° 55' W. 


83. 


14 a. 2 r. 9 p. 


100. 


26.88. 


122. 


103.6 miles. 


84. 


61 A. 2 R. 


103. 


6. 


124. 


33° 18' N. ; 


85. 


4 A. 2 r. 26 p. 


108. 


6. 




36° 24' W. 


86. 


13.93, 23.21, 32.5 ch.110. 


6087 feet. 


125. 


N. 28° 47' E. ; 


87. 


9 a. 


111. 


5° 25' S. ; 




1293 miles. 


89. 


876.31. 




457.5 miles. 


126. 


S. 50° 40' W. ; 


90. 


1229.5. 


112. 


460.8 ; 383.1 miles. 


250.8 ; 20° 9' W. 


92. 


1075.3. 


113. 


229 miles ; 


127. 


38°21'N. ; 


93. 


2660.45. 




lat. 11° 39' S. 




55° 12' W. 



12 TRIGONOMETRY. 



128. 171 miles ; 32° 44' W. 129. N. 36° 52' W. ; 36° 8' W. 

130. 173 miles ; 51° 16' S. ; 34° 13' E. 

131. S. 50° 58' E. ; 47° 15' N. ; 20° 49' W. 

132. N. 53° 20' E., 16° 7' W. ; or N. 53° 20' W, 25° 53' W. 
133. N. 47° 42.5' E., 19° 27' N. t 121° 51' E. ; or N. 47° 42.5' W., 19° 27' 

N., 116° 9' E. ; or S. 47° 42.5' E., 14° 33' N., 121° 48' E. ; or S 
47° 42.5' W, 14° 33' N, 116° 12' E. 

137. N. 73' k, 45 miles ; 42° 15' N., 69° 5' W. 

138. N. 72° W. f 287 miles ; 33° S.. 13° 2' E. 



ANSWERS. 13 



SPHERICAL TRIGONOMETRY. 

Exercise XIX. 

1. 110°, 100°, 80°. 2. 140°, 90°, 55°. 7. %ir,2n,™n. 

Exercise XX. 

3. (i.) Either a or b must be equal to 90°. (ii.) If a = 90°, then A = 90°, 
and B = b ; if b = 90°, then £=90°, and A = a. (iii.) c = 90°, 
J. = 90°, B=b. (iv.) c = 90°, A = 90°, B = 90°, (7= 90°. 

Exercise XXI. 

2. I. The cosine of the middle part = the product of the cotangents of 

the adjacent parts. 
II. The cosine of the middle part = the product of the sines of the 
opposite parts. 

Exercise XXII. 

24. A = 175° 57' 10", B = 135° 42' 50", (7= 135° 34' 7". 

25. C = 104° 41' 39", a = 104° 53' 2", b = 133° 39' 48". 

26. sin a = sin A esc C, cos b = — tan J. cot C, cos B = — sec 4 cos C. 

27. a = 60°, 6 = 90°, £=90°. 

28. The triangle is impossible ; why ? 

29. b = 130° 41' 42", c = 71° 27' 43", 4 = 112° 57' 2". 

30. a = 26° 3' 51", A = 35°, B = 65° 46' 7". 

31. Impossible; why? 

Exercise XXIII. 

1. cos A = cot a tan £ 6, sin \ B = esc a sin \ b, cos & = cos a sec £ b. 

2. sin £ 4. = £ sec £ a, or cos A = £ cos a sec 2 £ a, or tan £ 4 = sec a cos £ a. 

180° 180° 

3. sin £ A = sec J a cos , sin B = sin £ a esc , 

sin r = tan £ a cot 

n 

4. Tetrahedron, 70° 31' 43" ; octahedron, 109° 28' 18" ; icosahedron 

138° 11' 36" ; cube, 90° ; dodecahedron, 116° 33' 44". 

5. cot \ A = Vcos a. 



14 TRIGONOMETRY. 



Exercise XXV. 

1. (i.) tan m = tan b cos A, (ii.) tan m = tan a cos B, 

cos a = cos b sec m cos (c— m) ; cos 6 = cos a seem cos (c—m). 



Exercise XXVI. 

1. (i.) tan x = tan B cos c, (ii.) tan x = tan J. cos b, 

cos J. = cos B esc x sin (C— x) ; cos 5 = cos A esc x sin ((7— x). 



Exercise XXXI. 
4. 27.25". 

Exercise XXXII. 

1. If x denote the angle required, sin \x = cos 18° sec 9°, x = 148°42'. 

2. cos x = cos A cos 5. 

3. Let w = the inclination of the edge c to the plane of a and b. Then 

it is easily show v n that V= abc sin I sin w. Now, conceive a sphere 
constructed having for centre the vertex of the trihedral angle 
whose edges are a, b, c. The spherical triangle, whose vertices are 
the points where a, b, c meet the surface of this sphere, has for its 
sides I, m, n ; and w is equal to the perpendicular arc from the side 
I to the opposite vertex. Let L, M, N, denote the angles of this 
triangle. Then, by means of [38] and [47], we find that 
sin w = sin m sin N= 2 sin m sin ^iVcos J N 

= — — Vsin s sin (s — I) sin (s — m) sin (s — n), 
sinZ 

where s = %(l + m-\-n) ; 



hence, V= 2 abc Vsm s sin (s — I) sin (s — m) sin (s — n). 

4. (i.) 9,976,500 square miles ; (ii.) 13,316,600 square miles. 

5. Let m = longitude of point where the ship crosses the equator, b = 

her course at the equator, d = distance sailed. Then " 

tan m == sin ftan o, cos b = cos I sin a, cot d= cotl cos a. 

6. Let k = arc of the parallel between the places, x = difference re- 

quired ; then sin £ k = sin %dsecl. z = 90° ( v2 — 1). 

7. cos (m — m') = (cos d — sin Z sin V) sec Z sec V ; where m and w' are 

the longitudes of the places. 

9. 12 hrs. 42 min. 41 sec, very nearly. 10. 60°. 



ANSWERS. 15 



11 cos t = — tan d tan I ; time of sunrise = 12 o'clock a.m. ; time of 

t 15 

sunset = — o'clock p.m. ; cos a — sin d sec I. For longest day at 

15 
Boston : time of sunrise, 4 hrs. 26 min. 50 sec. a.m. ; time of sunset, 

7 hrs. 33 min. 10 sec. p.m. Azimuth of sun at these times, 57° 25' 
15" ; length of day, 15 hrs. 6 min. 20 sec. ; for shortest day, times 
of sunrise and sunset are 7 hrs. 33 min. 10 sec. a.m. and 4 hrs. 26 
min. 50 sec. p.m. ; azimuth of sun, 122° 34' 45" ; length of day, 

8 hrs. 53 min. 40 sec. 

12. The problem is impossible when cotc?<tan£; that is, for places in 

the frigid zone. 

13. For the northern hemisphere and positive declination, 

sin h = sin I sin d, cot a = cos I tan d. 
Example : h = 17° 14' 35", a = 73° 51' 34" E. 

14. The farther the place from the equator, the greater the sun's altitude 

at 6 a.m. in summer. At the equator it is 0°. At the north pole 
it is equal to the sun's declination. At a given place, the sun's 
altitude at 6 a.m. is a maximum on the longest day of the year, 
and then sin h = sin I sin e (where e = 23° 27'). 

15. cos t = cot I tan d. Times of bearing due east and due west are 

12 o'clock a.m., and — o'clock p.m., respectively. 

15 15 < r J 

Example : 6 hrs. 58 min. a.m. and 5 hrs. 2 min. p.m. 

16. "When the days and nights are equal, d= 0°, cost= 0, t = 90° ; that 

is, sun is everywhere due east at 6 a.m., and due west at 6 p.m. 
Since I and d must both be less than 90°, cos t cannot be negative, 
therefore, t cannot be greater than 90°. As d increases, t decreases ; 
that is, the times in question both approach noon. If Kd, then 
cos t > 1 ; therefore this case is impossible. If I = d, then cos t =1, 
and t = 0° ; that is, the times both coincide with noon. The ex- 

Elanation of this result is, that the sun at noon is in the zenith ; 
ence on the prime vertical. At the pole I — 90°, cos t = 0, 
t = 90° ; therefore the sun in summer always bears due east at 6 
a.m., and due west at 6 p.m. 

17. sin 1 = sidd esc h. 18. 11°50 , 35 ,/ . 

19. The bearing of the wall, reckoned from the north point of the hori- 

zon, is given by the equation cot a; = cos I tan d; whence, for the 
given case, x = 75° 12' 38". 

20. 55° 45' 6" N. 21. 63° 23' 41" N. or S. 

22. (i.) cos t = — tan I tan d ; (ii.) t = z ; (iii.) the result is indeterminate. 

23. cot a = cos I tan d. 28. sin d= sine sin v, tan r = cose tan v. 

25. h = 65° 37' 20". 29. d = 32° 24' 12", r = 301° 48' 17". 

26. h = 58° 25' 15", a =152° 28'. 30: d= 20° 48' 12". 

27. t = 45° 42', I = 67° 58' 54". 31. 3 hrs. 59 min. 27f sec. p.m. 
32. cos Ja=Vcos§(7 + h +p)cos%(l -f- /i— p)sec Zsec h, 



AXSTTERS. 







SURVEYING. 










Exercise I. 






1. 

2. 

3. 
4. 


8 a. 64 p. 
29 a. 7| p. 

4 A. 5^ P. 

U5& p. 


5. 3 a. 7^ p. 

6. 13 a. 6Jq p. 

7. 11 a. 157 p. 

8. 7.51925. 

EXERCISE II. 


9. 
10. 

11. 


13.0735. 

2 a :->i p, 
4 a. 35 "p. 


1. 

2. 

3. 


2 a. 26 p. 

:: a. 12 p. 

2 a. 54 p. 


5. B a. 54 p. 
E 5 a. 42 p. 
7. 2 A. 75 ?. . 


8. 
9. 

10. 


3 a. 122 p. 
6 a. 2 ?. 

9a.-: ?. 



A. 151 P. 



Exercise III. 
1. 2 a. 121 p. _ 98 a. 92 p. 



Exercise IV. 



8. 



AE=S.7b eh. 
AB=S.oO ch. ; 
.E<2 = 3.42 ch. 
AE=±.oo ch. 
AE=5.bO ch. 
CE = 4.455 ch. 
AD= 2.275 ch. ; 
BE =1.82 ch. 
.4Z> = 4.51 ch. ; 
5J7= 3.608 ch. 
The distances on i5 are 
2. 3. and 5 ch. 



9. .EJf(onJ£4)== 2.5087 ch. 
AlT((mAB)=6.439ch. 

10. Lr: EG > DF, 

f ^£7= 12.247 ch. 
,, \ AG = 9.798 ch. 

< AB = 8.659 ch. 
AF = 6.928 ch. 

11. Let DG>EF, 
OG = 14.862 ch. 
CD= 13.113 ch. 

CF= 11.404 ch. 
CE= 10.062 ch. 



then 



ANSWERS. 



17 



Exercise V. 

1. 9.5 ft. 

2. Third column : 26.944 opposite ; 25.286 opposite 4. 
Fifth column : 20, 19.5, 23, 22.3, 21.431, 20.4, 21.8, and 

24.1. 




O 1 2 3 4 5 6 

3. Column H.G. 20.8, 20.4, 20.0, 19.6, etc. 
Column G. 0.0, 5.3, 6.4, 7.4, 5.0, 5.1, etc. 




NAVIGATION. 

CHAPTEE I. 

DEFINITIONS. COURSE AND DISTANCE. 



§ 1. Definitions. 

Navigation is the art of conducting a ship from one port to 
another. 

The True Course of a ship is the angle which her path makes 
with the true meridian. 

The Correct Magnetic Course is the angle which the ship's 
path makes with the magnetic meridian. 

When a ship is steered on one course, her path crosses the 
meridians at the same angle. Such a line is called the Ship's 
Track, Ehumb-line, or Loxodromic Curve. 

The Distance is the path of the ship expressed in geographi- 
cal miles. 

The two most important problems of navigation are : 

1. To find the present position of the ship. 

2. To determine the future course. 

The position (latitude and longitude) of a ship at sea may- 
be found by two distinct methods, viz. : 

1. By* Dead-reckoning, which consists in keeping a record 
of the courses and distances sailed from a known point, and 
computing from these data the position arrived at. 

2. By Nautical Astronomy, which consists in determining 
position by observations of the heavenly bodies. 



NAVIGATION. 



When a ship is near land, the depth of water and the 
nature of the bottom, as determined by sounding, often serve 
to indicate her position. 

§ 2. The Mariner's Compass. 

When out of sight of land, vessels are steered by the Mari- 
ner's Compass, 

The Card of the compass consists of a circular plate of mica 
covered with paper, the circumference of which is usually 
divided into degrees. An inner circle is divided into thirty- 
two equal parts by radial lines called rhumb-lines. The ex- 
tremities of the rhumb-lines are called points, or rhumbs. 

Note. A figure of the card of the compass, and a table giving the 
angular distance of each point, half-point, and quarter-point from the 
meridian, will be found on page 55 of Wentworth & Hill's Tables. 

The principle upon which the points are named will be 
understood by inspecting a single quadrant. The point (N.E.) 
midway between N. and E. is named by writing these letters 
together. The point (N.N.E.) midway between N. and N.E. 
is named by joining these letters. The first point to the east 
of north is named N. by E., and the first point to the north of 
N.E. is named N.E. by N. In like manner, the points between 
E. and N.E. are named. 

Naming the points in order is called boxing the compass. 

One or more Magnetic Needles are attached to the lower sur- 
face of the card, parallel to the N. S. line, and having their 
N. and S. poles in the same direction as the corresponding 
points of the card. 

The card turns upon a pivot in the centre of a cylindrical 
metallic box called the Bowl, which is supported by Gimbals 
in a square box. 

In some compasses the card almost floats in alcohol, which 
relieves the pivot of the greater part of the weight of the card 
and needles. 

The Lubber-line is a distinct vertical line on the inner side 
of the bowl. 



DEVIATION. 



It is often convenient to refer to the points of the compass 
as right or left of north or south. Suppose an observer to 
stand at the centre of the card, facing north. Then all points 
between N. and E. are right of north (R. of N.), and all. points 
between N. and W. are left of north (L. of N.). In like man- 
ner, if the observer face south, all points between S. and W. 
are right of south (R. of S.), and all points between S. and E. 
are left of south (L. of S.). 

Thus, W.S.W. is 6 points, or 67° 30', R. of S. 

N.by W. is 1 point, or 11° 15', L. of N. 

S.E. is 4 points, or 45°, L. of S. 

E. by N. is 7 points, or 78° 45', R. of N. 

The compass is placed in a box called the Binnacle. When 
properly placed, the lubber-line is next to the bow of the ves- 
sel, and a line passing through the lubber-line and the centre 
of the card is parallel to the fore-and-aft line of the vessel. 

The Steering Compass is placed immediately before the helms- 
man. The Standard Compass is commonly placed near the 
middle of the vessel, in such a position as to be least affected 
by the iron used in the construction of the vessel. (See § 5.) 

§ 3. Variation of the Needle. 

This subject has been treated in " Surveying " (§ 23). The 
navigator is provided with charts which give the variation at 
different places. 

§ 4. Local Attraction. 

The disturbing effect upon the needle of iron outside of the 
ship (such as the iron in docks) is called local attraction. 

§ 5. Deviation. 

Deviation includes the changes in the direction of the needle 
caused by the iron in the ship. 

/Sub-permanent magnetism is the magnetism of the hard iron 
used in constructing the vessel. It depends upon the direc- 



NAVIGATION. 



tion of the ship's head while building and the direction of the 
dipping needle at the place of construction. 

Induced magnetism is the magnetism induced in the soft 
iron in the vessel by the earth's magnetic force. 

Retentive nno.gnetism is the temporary magnetism induced 
in an iron ship when her head is kept in one direction for 
some time. 

Semicircular Deviation is caused by sub-permanent magnetism 
and the transient induced magnetism of vertical masses of soft 
iron in the ship. In wooden ships it disappears when the 
ship's correct magnetic course is N. or S., and is greatest when 
the correct magnetic course is E. or W. In iron ships the 
neutral points are in the direction of the ship's head and stern 
while building. The part due to sub-permanent magnetism 
remains the same in kind, but is different in amount, in all 
latitudes, unless the ship be strained or subjected to mechani- 
cal violence. The part produced by induced magnetism in 
vertical iron changes in different latitudes with the inclination 
of the dipping needle. It disappears at the magnetic equator, 
and is of contrary names on opposite sides of this equator. 

Quadrantal Deviation is caused by the transient induced mag- 
netism of horizontal masses of soft iron in the ship. It disap- 
pears when the correct magnetic course is N., S., E., or W., 
and is greatest when the correct magnetic course is N.E., N.W., 
S.E., or S.W. It remains the same in all magnetic latitudes, 
and is not appreciably affected by the lapse of time. 

The amount of deviation may be found as follows : 

1. By the "known bearing of a distant object on shore. The 
ship is swung around so as to bring its head successively to the 
thirty-two points of the compass, and the bearing of the object 
on shore, as given by the standard compass when the ship's 
head is steadied on each point, is recorded. The difference 
between the known correct magnetic bearing of the object 
and the successive bearings given by the standard compass 
will give the deviation for these points. 



DEVIATION. 



2. By reciprocal bearings. If there is no suitable object on 
land, a second compass may be placed on shore, and the mu- 
tual bearings of this instrument and the standard compass on 
shipboard, when the ship's head is steadied on a certain point, 
taken at the same instant by preconcerted signal. The bear- 
ing given by the compass on shore is then reversed. The 
difference between this reversed bearing and the bearing given 
by the standard compass will be the deviation for this point of 
the compass. In like manner, the deviation for each point 
may be found. 

3. By amplitudes and azimuths of a heavenly body. (§ 32.) 

The deviation is named as follows : If the correct magnetic 
bearing of the distant object (or the reversed bearing of the 
shore compass) is to the right of the reading of the standard 
compass, the deviation is easterly ; and vice versa. 

A Deviation Table, or deviation card, is a tabulated statement 
of the amount of deviation for each point of the compass, when 
the ship is on an even keel. 

The Heeling Error must be estimated and allowed for separ- 
ately. 

Navigators commonly attempt to compensate for deviation 
by placing magnets and pieces of soft iron near the compass in 
such positions as to overcome the deviation. 

Example 1. A ship is headed N.E. A distant object on 
shore bears S. 70° W., by the standard compass. The correct 
magnetic bearing of the object from the ship is known to be 
S. 89° W. Find the deviation. 

Correct magnetic bearing S. 89° W., or 89° R. of S. 

Observed bearing S. 70° W., or 70° R. of S. 

Deviation 19° E. 

The deviation is east, because 89° R. of S. is to the right of 
70° R. ofS. 



6 NAVIGATION. 



Example 2. A ship is headed E. The hearing of the com- 
pass on shore from the ship, as given by the standard compass, 
is S. 21° W. The standard compass on the ship bears N. 8° W. 
from the - compass on shore. Find the deviation. 

Correct magnetic bearing (shore reversed) . S. 8°E., or 8° L. ofS. 
Observed bearing S. 21° W., or 21° R. of S. 

Deviation 29° W. 

The deviation is west, because 8° L. of S. is to the left of 
21° R. of S. 

§ 6. The Wind. 

The wind is named from the point of the compass from which 
it blows. 

"Windward is the direction from which the wind blows. 

Leeward is the direction toward which the wind blows. 

"When the wind is directly behind a ship it is said to be 
right aft, or right astern, The ship is then sixteen points from 
the wind. 

A ship is said to be on the starboard tack or the port tack 
according as the wind blows on the right or left side of the 
ship. 

When the wind blows at right angles to the side of a ship, 
the ship is said to have the wind on her beam, The ship is 
then eight points from the wind. 

When a ship is more than eight points from the wind, the 
wind is said to be abaft the beam ; when the ship is less than 
eight points from the wind, the wind is said to be before the 
beam. 

A ship is said to be close-hauled when she is as near the 
wind as she will lie ; that is, when she is headed as near as 
possible toward the point from which the wind blows. Few 
ships can lie closer than five points from the wind. 



LEEWAY. 




§ 7. Leeway. 

The angle which the fore-and-aft line of a vessel makes with 
her track is called leeway. 

Let the arrow (Fig. 1) represent the direction of the wind, 
and CB the direction of the ship. 
Let CB represent the distance the 
ship would go in a given time if she 
were not allowed to move sideways, 
and CD the distance she would go in 
the same time if she were not allowed 
headway. By a principle of mechan- 
ics, the ship would, in the given time, F,g ' '" 
describe CA the diagonal of the parallelogram of which CB 
and CD are adjacent sides. 

The angle BCA is the leeway. 

The amount of leeway depends upon the trim of the ship, 
her velocity through the water, the sails set, etc. 



§ 8. Reducing the Courses. 

The Compass Course is the course steered. 

The true course is equal to the compass course corrected for 
variation, deviation, and leeway. 

From the preceding explanations, the following rules for 
changing from one course to the other are evident : 

To find the true course from the compass course : 

Variation :{ Allow easterlv variation to the right. 
I Allow westerly variation to the left. 

j, • , • .f Allow easterly deviation to the right. 
I Allow westerly deviation to the left. 

T f Allow leeway to the left on the starboard tack. 

I Allow leeway to the right on the port tack. 



8 NAVIGATION. 

To find the compass course from the true course : 

Variation • i ^ ow eas "terly variation to the left. . 
I Allow westerly variation to the right. 

Deviatio • I -^ ow eas t e rly deviation to the left. 
I Allow westerly deviation to the right. 

Leeivav • I ^ ow l eewa J to tne r ig nt on the starboard tack. 
I Allow leeway to the left on the port tack. 



Example 1. The compass course is E.N.E., wind S.E., 
leeway 3 points, variation 10° E., deviation 2° 30' E. Find 
the true course. 

Since the wind is S.E., and the compass course E.N.E., the ship is 
on the starboard tack ; hence, leeway is allowed to the left. 

Compass course 6 pts. R. of N. 

Leeway 3 pts. L. 

The compass course corrected for leeway 3 pts. R. of N. 

= 33°45 / R.ofN. 

Variation and deviation (10° E. + 2° 3C E.) 12° 3C/ R. 

True course 46° 15 7 R. of N. 

= N. 46° lc/ E. 

Example 2. The compass course is W. S.W., wind S., lee- 
way 2^ points, variation 30° E., deviation 18° "W. Find the 
true course. 

Compass course 6 pts. R. of S. 

Leeway 2J pts. R. 

8£ pts. R. of S. 
= 7£ pts. L. of N. 
= 84°23'L. ofN. 
Variation and deviation (30° E. - 18° W. = 12° E.) 12° tf R. 

True course 72° 23' L. of N. 

= N. 72° 23' Y\\ 

Example 3. The true course is N.E. by N., the vessel is on 
the port tack, leeway 10°, variation 1 point W., deviation 5° 30' 
E. Find the compass course. 



THE LOG. 



9 



True course 33° 45> R. of N. 

Variation 11° 15' R. 

Correct magnetic course 45° / R. of N. 

Leeway ' 10° 0' L. 

35° (yR.ofN. 
Deviation (for 35° R. of N. = N.E. by N. nearly) . . 5° 3C L. 

Compass course 29° 30' R. of N. 

= N. 29° 30 7 E. 



Exercise I. 





Compass 
Course. 


Wind. 


Leeway. 


Variation. 


Deviation. 


True Course. 


1. 


s. 


E.S.E. 


1\ pts. 


52°0'W. 


2° / E. 


Required. 


2. 


W.N.W. 


N. 


3 " 


42° (X E. 


18°30 / W. 


Required. 


3. 


S.S.E.JE. 


S.W.JS. 


3* " 


2\ pts. E. 


11 pts. W. 


Required. 


4. 


Required. 


S. by W. 


I'- 


10° 30' E. 


19° O'W. 


S.79°W. 


5. 


W.JN. 


N.N.W. 


ll" 


8° 3(y E. 


15° 35' E. 


Required. 


6. 


E.iN. 


N.N.E. 


2\ « 


13° O'W. 


20° O'E. 


Required. 


7. 


Required. 


N. by W. 


* " 


14° (/E. 


19° O'E. 


S. 85° E. 


8. 


W. 


N.N.W. 


It" 


18° 30' E. 


21° O'W. 


Required. 


9. 


E.JS. 


N.N.E. 1 E. 


2£» 


21° O'E. 


4° O'W. 


Required. 


10. 


Required. 


N. by W. 


2|" 


2 pts. W. 


Si pts. E. 


E.byS.lS. 


11. 


Required. 


N.E. 


3} " 


2| pts. E. 


1$ pts. E. 


N. by W. 


12. 


Required. 


S.S.W. 


2*" 


2| pts. E. 


| pts. E. 


N.N.W. 


13. 


Required. 




" 


7° 0' W. 


15° 0' W. 


S. 64° E. 


14. 


Required. 




" 


6° C/E. 


20° O'W. 


N.44°W. 


15. 


N.65°W. 




" 


10° 0' E. 


3° O'E. 


Required. 


16. 


S. 15° W. 




" 


6° O'W. 


18° O'E. 


Required. 


17. 


S. 18° E. 




" 


25° 0' E. 


10° V E. 


Required. 


18. 


N. 30° E. 


S. by W. 


11 " 


12°0 / E. 


10° 0' W. 


Required. 



§ 9. The Log. 



The rate at which a vessel moves through the water is com- 
monly determined by the Log. 

The Common Log consists of the log-chip (often called simply 
the log), the log-line, and the sand-glass, 



10 



NAVIGATION. 




Fig. 2. 



The Log-chip (Fig. 2) is a flat piece of wood in the form of a 
sector, of about five inches radius, 
having its arc loaded with lead to 
make it float in an upright position. 
The Log-line, to which the log-chip 
is attached by three strings, is an 
ordinary cord about 150 fathoms 
long. The stray-line is the part of the log-line next the log- 
chip, and is commonly about two-thirds the length of the ship. 
The end of the stray-line is marked by a rag tied into the line. 
The remainder of the line is divided into equal parts called 
knots, and each knot is divided into tenths. Beginning at the 
end of the stray-line, the end of the first knot is marked by a 
piece of cord having one knot tied in it; the end of the second 
knot is marked by a cord with two knots tied in it ; the end 
of the third knot is marked by a cord with three knots tied 
in it, etc. 

A 30-seconds, 28-seconds, or 14-seconds sand-glass (Fig. 3) 
may be used. 

The knot is made the same part of a 
geographical mile (about 6080 ft.) that the 
time required for the sand to run through 
the glass is of an hour. The knot corre- 
sponding to a 30-seconds glass would be 
determined by the following proportion : 




3600 s. : 30 s. : : 6080 ft. : x = 50* ft. 

As a vessel generally overruns her 
reckoning, the knot corresponding to a 
30-seconds glass is made 50 ft. The pro- 
portion would give 47-^j ft. as the length 
Fi e- 3 - of the knot corresponding to a 28-seconds 

glass, but in practice it is made 46 ft. 8 in. 
The log-line is wound upon a reel. (Fig. 4.) 
The operation of heaving the log is as follows : The officer 
throws the log-chip into the water on the lee side of the vessel. 



THE LOG. 



11 




Fig. 4. 



It remains nearly stationary in the water while the line is 
reeled off as the 
vessel moves for- 
ward. The in- 
stant that the 
mark indicating 
the end of the 
stray-line passes 
over the side of 
the ship, the glass 
is turned. The 
operator then counts the number of knots run out during the 
time required for the sand to run through the glass. The line 
is given a quick jerk, which detaches the two lower of the 
three cords by which the log-chip is attached to the line, the 
log falls flat, and is readily drawn in. 

Since each knot is (approximately) the same part of a geo- 
graphical mile that the time is of one hour, the number of 
knots run out, while the sand is running through the glass, 
indicates the rate of the vessel in geographical miles per hour. 

The Patent Harpoon Log, one form of which is shown in Fig. 
5, consists of a brass cylinder, which is towed through the 




Fig. 5. 

water after the vessel by a line about fifty fathoms long. The 
cylinder consists of two parts. The hinder part, or rotator, is 
made to revolve by four curved blades as the log is drawn 
through the water ; the forward part, which is kept from 
revolving by two straight blades, contains a registering appara- 
tus, which is connected with the rotator, and shows the dis- 
tance. 



12 



NAVIGATION. 



The Patent Taffrail Log, one form of which is shown in Fig. 6, 
consists of a registering apparatus which is permanently at- 
tached to the taffrail of the vessel, and a rotator which is towed 
through the water by a cord connected with the register. The 
revolutions of the rotator are communicated to the register by 
the cord. 




Fig. 6. 



The Ground Log consists of the ordinary log-line with a hand- 
lead substituted for the log-chip. "When it is hove, the lead 
sinks to the bottom and remains stationary. The line reeled 
off gives the speed over the bottom. This log is not affected 
by tides or currents. It may be used in shallow water. 

The rate of paddle-steamers can be estimated from the num- 
ber of revolutions of the paddle-wheels, quite as accurately as 
by the best patent log, especially in smooth water. 



The Different Logs Compared. 

The best results are obtained by the use of the harpoon log, 
if it be in good order ; but the rapid motion of parts of the 
registering apparatus soon puts the best patent log out of 



CURRENTS. 13 



order. Hence, fast ocean steamers ordinarily use the common 
log (hove every two hours) when in the open sea, and both 
the common and the patent log when nearing land. 

The taffrail log is the most convenient, since the registering 
apparatus can be examined at any time without taking the 
rotator from the water ; and if the rotator be lost it may be 
replaced at small cost, whereas, if the harpoon log be lost (a 
not infrequent occurrence), the cost of replacing it is consider- 
able. But the taffrail log is not considered as reliable as the 
harpoon log, although it is growing in favor. 

§ 10. Currents. 

The set of a current is its direction, or the point of the com- 
pass toward which it is moving. 

The drift of a current is its rate in miles per hour. 

The method of allowing for the effect of a current upon a 
ship will be explained hereafter (§18). 



CHAPTER II. 

THE SAILINGS. 



§ 11. Definitions. 

The Difference of Latitude of two places is the arc of a 
meridian included between their parallels. 

The Middle Latitude of two places is the latitude of a par- 
allel midway between them. 

The Difference of Longitude of two places is the arc of the 
equator included between their meridians. 

The Departure is the distance between the meridian left and 
the meridian arrived at, measured on a parallel. 



§ 12. Notation. 

C — the true course. 

D = the distance in geographical miles. 

L = latitude in general. 

V = the latitude left ("latitude from"). 

L n = the latitude arrived at ("latitude in"). 

L m — the middle latitude. 

L d = the difference of latitude. 



A 
A' 
A" 

K 


= longitude in general. 

= the longitude left ("longitude from"). 

= the longitude arrived at ("longitude in"). 

= the difference of longitude, 

= the departure. 



PLANE SAILING. 



15 



§ 13. Plane Sailing. 

Case I. When the distance is so small that the curvature of 
the earth may be neglected. 

Let CA (Fig. V) be the distance sailed, CB the meridian 
through C, and AB the parallel through A. 
Then, angle ACB = C, 
CA = D, 
CB = L d , 
AB=p. 

By Plane Trigonometry, 
sin (7=£, 

p — D sin C, 



008 ff=^, 

L d = D cos C, 
L d 



sinC 



D 



cos 



tan C= 3L ; 
D d 
p = L d t&iiC', 

£ — P 

tanC 




Case II. When the distance is so great that the curvature of 
the earth cannot be neglected. 

Let PCD (Fig. 8) represent a por- 
tion of the earth's surface, P the 
pole, CA the ship's track between 
C and A, PC and PD the meri- 
dians, and CD and BA the par- 
allels through (7 and A. 

Let CA be divided into parts Ch, 
hj, etc., so small that the curvature 
of the earth may be neglected for 
these distances ; and let gh, ij, etc., 
be intercepts on the parallels corresponding to Ch, hj, etc. 

Angle PCA =* PhA = PjA = = C. 

By Plane Trigonometry, 

gh = Ch sin C, 
ij = hj sin C, 




gh + ij + = (0h + hj+ ) sin C. 



16 



NAVIGATION. 



Hence, 
Also, 



p = D sin C. 
Cg= Ch cos C, 
hi = hj cos C, 



(a) 



.-. (7# + hi + = (Ch + hj -f ) cos (7. 



Hence, 

From (a) and (b), 

sinO- P 



D cos C. 



D' 

p = D sin 

sinU 



cos () = ■=*, 



[1] 



D 



D 
DcosO, 



cosO 



[2] 



tanO = ^-, 
p = LrftanC, 
I/- P . 



ib) 



[3] 



These formulas are the same as in Case I. ; but in Case II. 
p is evidently greater than BA and less than CD. (See § 15.) 

Plane Sailing, as here treated, may be denned as the method 
of establishing the relations between course, distance, differ- 
ence of latitude, and departure, on the supposition that small 
portions of the earth's surface may be regarded as plane. 

The several problems are readily solved by formulas [1], 
[2], and [3]. 

In practice, the method by inspection is commonly used by 
navigators. A number of examples will be solved by this 
method in order to illustrate the use of the Traverse Tables. 

Example 1. A ship sails from latitude 40°20'N., on a 
N.N.E. course for 92 miles. Find the departure and latitude in. 

Solution I. By computation. The true course N.N.E. = N. 22° 3CX E. 
To find the departure ■ 

p = D sin 0. [1] 

D = 92. I logD =1.96379 

C = 22° 3(7 | log sin = 9.58284 

logp = 1.54663 
:.p = 35.2 mi. 



To find the difference of latitude : 


L,= 


DcosO. [2] 


logD 


= 1.96379 


log cos 


= 9.96562 


log id 


= 1.92941 




-. Zrf = 85.0mi. 




= 1°25 / N. 




1/ = 40° 2<y N. 



PLANE SAILING. 17 



Solution II. By inspection. Since the hypotenuse and an acute angle 
of a right triangle are given to find the sides, the Traverse Table may be 
used. Corresponding to the course N.N.E. (2 points), and distance 92 
miles, Table VIII.* gives latitude = 85, and departure = 35.2. 

Example 2. A ship sails S.W. by W. 488 miles. Find the 
difference of latitude and departure. 

By inspection. S.W. by W. = 5 points. In Table VIII., at the bottom 
of page 73, we find 5 points ; hence, we look at the bottom of the page 
for the designations of the columns. The table only extends to a distance 
of 300 ; hence, we find the latitudes and departures corresponding to the 
distances 300 and 188, and take their sum. 

For distance 300 latitude = 166.7 departure = 249.4 

" 188 " = 104.4 " = 156.3 

" 488 " =271.1 . " =405.7 

Or, we may divide the given distance by any convenient number, and 

find the latitude and departure corresponding to the quotient (the course 

remaining unchanged), then multiply each of these by the divisor of the 

distance. 

488 -*- 2 = 244. 

For distance 244 latitude = 135.6 departure = 202.9 

2 2 2 

488 271.2 405.8 

The results differ slightly by the two methods; but this 
small discrepancy is of no importance in practice. (See § 18.) 

Example 3. Course 6J points, distance 21.7 ; required the 
difference of latitude and departure. 

By inspection. Q\ points is found at the bottom of page 67, Table VIII. 
Corresponding to distance 217 (21.7 X 10) the difference of latitude is 
63.0, and the departure 207.7. Dividing each of these numbers by 10, the 
required difference of latitude is found to be 6.3, and the departure 20.8. 

Example 4. If a ship run S.E. by S. from 1° 44' north lat- 
itude, and is then by observation in 2° 46' south latitude, 
what are the distance and departure ? 
S.E. by S. = 3 points. 1° 44' N. 
2° 46' S . 
4° 30' = 270 mi. = difference of latitude. 

* Wentworth & Hill's Tables. 



18 



NAVIGATION. 



On page 73, Table VIII. .we find 3 points at the top ; but the greatest 
difference of latitude on this page is 249 ; hence, we divide the given dif- 
ference of latitude by 2. 270 -=- 2 = 135. The nearest difference of lati- 
tude to 135 is 134.7, opposite which the distance is 162, and the departure 
90. Multiplying each of these numbers by 2, the required distance is 324 
(= 162 X 2), and the departure 180 (= 90 x 2). 

Example 5. A ship sails N. 25° W. until the departure is 
98 miles. Find the distance and difference of latitude. 

Since the course is given in degrees, we turn to Table IX. At the top 
of page 102 we find 25° , and opposite departure 98, we find the required 
distance 232, and difference of latitude 210.3. 



Example 6. Find the course and distance corresponding 
to a difference of latitude of 696 miles, and a departure of 186 
miles, the course lying between N. and E. 



Solution I. By computation. 
To find the course : 



P 



186 



tan0 = £ 
±* d 

logp 
colog Ld -- 



[3] 



2.26951 
7.15739 



log tan (7=9.42690 

.-. C7=N.14°5^E. 



To find the 


distance : 


D = 


P 
sinO 




logp = 
colog sin C= 


2.26951 

0.58795 


log D 


:D = 


2.85746 
720.2 



[1] 



Solution II. By inspection. Since the given difference of latitude 
exceeds the limits of the tables, we divide each of the given numbers by 
3 ; 696 -*- 3 = 232, 1S6 h- 3 = 62. If the course be desired in degrees, we 
now turn to Table IX. and look for a difference of latitude of 232 oppo- 
site a departure of 62. On page 92, the nearest approximation is found. 
At the top of this page the angle is 15° ; hence, the required course is 
N. 15° E. If the course be desired in points, we turn to Table VIII. The 
nearest approximation is found on page 66, at the top of which stands 
1} points. Hence, the required course is N. by E. \ E. The distance is 
720 (= 240 x 3) miles. 

Note. The answers, given at the end of this volume, to the exercises 
of this chapter, are obtained by computation. The student should, how- 
ever, solve both by computation and inspection. 



PARALLEL SAILING. 



19 



Exercise II. 



1 


U. 


L". 


C. 


D. 


P- 


1. 


49° 57' N. 


Required. 


S.W. by W. 


488.0 


Required. 


2. 


1° 45' N. 


Required. 


S.E. by E. 


487.8 


Required. 


3. 


3° 15> S. 


Required. 


N.E. by E. f E. 


449.1 


Required. 


4. 


2° 10' S. 


Required. 


N. by E. 


267.0 


Required. 


5. 


41° 3V N. 


Required. 


SOT. 


295.5 


Required. 


6. 


21° 59' S. 


24° 49' S. 


Required. 


360.0 


Required. 


7. 


2° 9'S. 


3° 11' N. 


Required. 


354.0 


Required. 


8. 


1° 30' N. 


0° 26' S. 


S. by W. 


Required. 


Required. 


9. 


40° 17' N. 


37° 6'N. 


S. by W. J W. 


Required. 


Required. 


10. 


38° O'N. 


Required. 


S.W. by W. 


Required. 


48.2 


11. 


18° 25' N. 


Required. 


S.W.byW.fW. 


Required. 


65.1 


12. 


50° 18' N. 


54° 48' N. 


Required. 


299.0 


Required. 


13. 


32° 30' N. 


19° 59/ N. 


Required. 


812.0 


Required. 


14. 


2° 8'S. 


Required. 


N. 11° E. 


500.0 


Required. 


15. 


20° 21' S. 


Required. 


N. 20° E. 


402.0 


Required. 


16. 


40° 25' S. 


Required. 


N. 87° E. 


240.0 


Required. 


17. 


20° 48' N. 


17° 13' N. 


Required. 


Required. 


289.2 W. 


18. 


51° 45' N. 


53° 11' N. 


Required. 


Required. 


128.0 E. 


19. 


0° 2V S. 


0° 18' N. 


Required. 


Required. 


142.7 E. 


20. 


40° 20' N. 


41° 37' N. 


Required. 


Required. 


52.6 W. 



§ 14. Parallel Sailing. 



Let P (Fig. 9) represent the pole 
earth, A and B two places on the same 
parallel AM, PAE and PBQ meri- 
dians through A and B intercepting 
EQ on the equator. Let AD and BB 
be perpendicular to PO. Join AO. 
AB is the departure or meridian dis- 
tance. Kegard the earth as spherical. 
AE= BQ = EOA = OAB 
= latitude of A and B. 
In the right triangle AOB, 

AB AB 



O the centre 



cosOAD=^ = 



AB 



AO B 
B cos L. 



or cosX 




20 



NAVIGATION. 



By Geometry, 



or 



EQ 



And 



AB: 

P • 

K = 



EO 
R 



AB 
R cos L. 

\ d cos L. 1 
p sec L. J 



[4] 



Since p=X d cos L, X d is the hypotenuse of the right triangle 
of which L is an acute angle and p the side 
adjacent, as in Fig. 10. 

Since p = X a cos L, the meridian distance 
varies as the cosine of the latitude. 

Let X d = 1° = 60' = 60 miles, then the 
above formula becomes^* = 60 cos L. 

Hence, to find the number of geographi- 
cal miles in 1° of longitude at any place, 
multiply 60 by the cosine of the latitude. 

The number of statute miles in a degree of longitude may 
be found by multiplying the numbers in the table by • § ^ i , or 
by using the formula^ = 69.16 cosX. 

Table showing the Number of Geogeaphical Miles in a Degree 
of Longitude in every Latitude. 




L 


Miles. 


L. 


Miles. 


L 


Miles. 


L. 


Miles. 


L. 


Miles. 


L 


Miles. 


1° 


59.99 


16° 


57.67 


31° 


51.43 


46° 


41.68 


61° 


29.09 


76° 


14.51 


2° 


59.96 


17° 


57.38 


32° 


50.88 


47° 


40.92 


62° 


28.17 


77° 


13.50 


3° 


59.92 


18° 


57.06 


33° 


50.32 


48° 


40.15 


63° 


27.24 


78° 


12.48 


4° 


59.85 


19° 


56.73 


34° 


49.74 


49° 


39.36 


64° 


26.30 


79° 


11.45 


5° 


59.77 


20° 


56.38 


35° 


49.15 


50° 


38.57 


65° 


25.36 


80° 


10.42 


6° 


59.67 


21° 


56.01 


36° 


48.54 


51° 


37.76 


66° 


24.41 


81° 


9.38 


7° 


59.56 


22° 


55.63 


37° 


47.92 


52° 


36.94 


67° 


23.44 


82° 


8.35 


8° 


59.42 


23° 


55.23 


38° 


47.28 


53° 


36.11 


68° 


22.48 


83° 


7.32 


9° 


59.26 


24° 


54.81 


39° 


46.63 


54° 


35.27 


69° 


21.50 


84° 


6.28 


10° 


59.09 


25° 


54.38 


40° 


45.96 


55° 


34.41 


70° 


20.52 


85° 


5.23 


11° 


58.89 


26° 


53.93 


41° 


45.28 


56° 


33.55 


71° 


19.53 


86° 


4.18 


12° 


58.69 


27° 


53.46 


42° 


44.59 


57° 


32.68 


72° 


18.54 


87° 


3.14 


13° 


58.46 


28° 


52.97 


43° 


43.88 


58° 


31.79 


73° 


17.54 


88° 


2.09 


14° 


58.22 


29° 


52.47 


44° 


43.16 


59° 


30.90 


74° 


16.54 


89° 


1.05 


15° 


57.95 


30° 


51.96 


45° 


42.43 


60° 


30.00 


75° 


15.53 


90° 


0.00 



PARALLEL SAILING. 



21 



Example 1. A ship in latitude 42° changes her longitude 
3° 20' by sailing on this parallel. Find her departure. 



Solution I. 

p = X d cos L. 

p = 60x3^xcos42 c 
= 60 x H X .7431 
= 148.6 miles. 



[4] 



Solution II. 
for L = 42°, 



The table gives 



1° 



of A = 44.59. 
of A = 44.59 x 3 1 
= 148.6 miles. 



Solution III. By referring to Fig. 10, it will be seen that the given 
latitude corresponds to course, and difference of longitude to distance. 
Hence, p is the latitude in the Traverse Table corresponding to course 
42° and distance 60 X 3£ = 200, which gives 148.6 miles, as before. 

Example 2. A ship sails 176.2 miles due west from the 
Lizard, in latitude 49° 57' N., and longitude 5° 14' W. Find 
the longitude in. 

Solution I. 



K 



_ p 



p = 176.2 
L = 49° 57' 

A' = 5° 14' W. 
A d =4°34 , W. 

a"= 9° 48' W. 



cosL 

\ogp = 2.24601 

colog cos L = 0.19148 

logA d 
.'.Ad 



M 



2.43749 

273.8 mi. 

4°34'W. 



Solution II. 

From the table, 

1° A for 49° L = 39.36 
1° A for 50° L = 38.57 

By interpolation, 
1° A for 49° 57 7 L = 38.61 
176.2-5-38.61= 4.56 
A d = 4.56° = 4° 34'. 



Exercise III. 





L 


P- 


y. 


A' . 


1. 


55° 55' 


Required. 


2° IV W. 


12° 52' E. 


2. 


52°0(y 


Required. 


0° 59' W. 


2° 24' E. 


3. 


61° 25' 


Required. 


179° 20' W. 


176° 52' E. 


4. 


56°0(y 


Required. 


3° 12' W. 


4° 8'E. 


5. 


80° OO 7 


Required. 


10° 0'W. 


17° 41' W. 


6. 


60° 00' 


204.0 E. 


160° 2'E. 


Required. 


7. 


51° 28' 


70.9 E. 


32° 7'W. 


Required. 


8. 


64° 16' 


265.7 W. 


170° 0'W. 


Required. 


9. 


46° 37' 


352.0 E. 


163° 42' E. 


Required. 


10. 


39° 57' 


398.0 W. 


4° 8'W. 


Required. 



22 



NAVIGATION. 



11. From latitude 32° 3' S., longitude 179° 45' W., a ship 
makes 54 miles west (true). Required the longitude in. 

Note. In changing from the trne to the compass course, apply varia- 
tion and leeway, then deviation, which is supposed to be given for the 
course to which it is directly applied. 

12. From latitude 35° 30' S., longitude 27°28'W., a ship 
sails east (true) 301 miles. Required the longitude in and 
the compass course ; variation If points E., leeway j point to 
the left, deviation 8° 50' E. 



§ 15. Middle Latitude Sailing. 

Middle Latitude Sailing is a combination of plane and par- 
allel sailings, chiefly for the purpose of determining difference 
of longitude. 

Let CA (Fig. 11) represent the ship's track, CD and BA 
paraUels, and PBC and PAD 
meridians through C and A respec- 
tively. Let EFhe a parallel mid- 
way between Cand A. 

It is shown in plane sailing that 
p is. less than CD and greater than 
DA. It is nearly equal to EF, 
and exactly equal to an intercept 
a little nearer to the pole than 
EF. 

Hence, D m = J(i'+Z"), nearly, 
if both places are on the same side of the equator. 

In middle latitude sailing p is first found as in plane sail- 
ing, then X d is found, on the supposition that X d is equal to 
the change of longitude which the ship would have made by 
sailing a distance p on a parallel in latitude L m . 
By parallel sailing, 




p = X d cos L m 
By plane sailing, 

p = D sin C, 



p sec L r 



tanC=-f-. 



[5] 



MIDDLE LATITUDE SAILING. 



23 



Hence, by combining, 
tanO 



K cos L m 
D sin sec L„ 



[6] 

[7] 

The formula L m = J (X'-f- B") may be used without sensible 
error in low latitudes when the course is not less than about 
45°, and both places are on the same side of the equator. 

When the places are on different sides of the equator, and 
the distance not great, p = X d may be used ; but when the 
distance is considerable, the following method is preferable. 

Let CA (Fig. 12), represent the rhumb-line crossing the 
equator EQ at B. Let CE and 
AQ represent meridians. 

For the rhumb CB the middle 
latitude is I CE; hence, EB (the 
difference of longitude of C and 
B) may be found. For the 
rhumb BA the middle latitude 
isUC; hence, BQ (the differ- 
ence of longitude of B and A) 
may be found. 




Fig. 12. 



Then 



\ d = EB+BQ. 



In high latitudes, or in any latitude, if the course be less 
than about 45° and the distance great, E 

£ ra = *(•#+ £")+<*, 

d being a small arc. This correction of the 
middle latitude is seldom used in practice. 

The relations of the elements of middle 
latitude sailing may be represented by com- 
bining the triangles of plane and parallel 
sailings, as in Fig. 13. 



ACB=C, 
CA = B, 
CB = L dt 



AB=p, 

AE=\ d , 

BAE=L m 




Fig. 13. 



24 



NAVIGATION. 






Example 1. A ship sails from Sandy Hook light, in lati- 
tude 40° 28' N., longitude 74° W., on an E.S.E. course, 62 
miles. Find the latitude and longitude in. 

Solution I. By computation. 



[2] 
1.79239 
9.58284 



To find the latitude in : 

L d = D cos 0. 

D = 62, logD 

C = 67°30', log cos 

log L d =1.37523 
.-. Z d = 23.7 = 24'S. 

V = 40° 28' N. 
L d = 24' S. 

£" = 40° 4'N. 

To find the middle latitude : 
L' = 40°28'N. 
Z" = 40° 4'N. 



To find the longitude in : 
\ d = D sin sec L, 
logD =1.79239 
log sin =9.96562 
log sec Z m = 0.11745 

log A* 

/. A d 



m 



1.87546 

75.1 

1° 15' E. 



a' = 74° (y W. 
\ d = 1° 15' E. 



A" = 72° 45' W. 



2 )80° 32' 
£ m = 40°16' 

Solution II. By inspection. In Fig. 13, it is evident that if we enter 
the Traverse Table with the course and distance, the difference'of latitude 
and departure may be found as in plane sailing, p. 17, Ex. 1. Corre- 
sponding to the course E.S.E. and distance 62, Table VIII., page 69, gives 
the difference of latitude 23.7, and the departure 57.3 ; whence, the lati- 
tude in and middle latitude may be found as in Solution I. 

In Fig. 13, the difference of longitude AE is the distance corresponding 
to the course E and the departure AB. But the angle E= 90° — BAE; 
that is, the angle E is the complement of the middle latitude. In this 
Example, the middle latitude is 40° 16' ; whence, the co-middle latitude 
is 90° -40° 16' =49° 44' =50° approximately. Corresponding to the 
course 50° and the departure 57.3 (the nearest is 57.5), Table IX., page 
117, gives the distance 75, which is the difference of longitude required ; 
whence, the longitude in may be found as in Solution I. 

Example 2. A ship sails from latitude 20° N., longitude 
160° W. until the difference of latitude is 1° 50' S., and the 
departure 440.2 miles W. Find the latitude in, the longitude 
in, the course, and distance, by inspection. 



MIDDLE LATITUDE SAILING. 



25 



1/ =20° C N. 


U = 20° (y N. 


A' = 160° O 7 W. 


L d = i°5(ys. 


£" = 18° 10' N. 


A d = 7° 45' W. 


±J' = 18° 10' N. 


2)38° 10' 

Z m =19° 5' 
coZ w =70°55 / 


a" = 167° 45> W. 



To find the longitude in. Since 440.2 exceeds the limits of the table, 
one-third of this number is taken. 440.2 -=- 3 = 146.7. Corresponding 
to course 71° (= co L m ) and departure 146.7, Table XI., page 96, gives 
the distance 155. 155 X 3 = 465 = 7° 45', which is the difference of lon- 
gitude ; whence, the longitude in is found to be 167° 45' W. 

To find the course and distance. Since the departure 440.2 exceeds the 
difference of latitude 110 (=1° 50'), the course and designations of the 
columns must be sought at the bottom of the page. 110-=- 2 = 55, and 
440.2 -=- 2 = 220.1. Corresponding to the difference of latitude 55 and the 
departure 220.1 (the nearest are 54.9 and 220.3), Table XI., page 91, 
gives the distance 227. 227 X 2 = 454. At the bottom of the page, the 
course 76° is found. Hence, the course and distance required are S. 76° W. 
454 miles. 

Example 3. A navigator wishes to sail from the Lizard, 
in latitude 49° 57' N., longitude 5° 14' W., to St. Mary's 
Island, in latitude 37° N. and longitude 25° 6' W. Find the 
true course and distance. 



To find L d : 
U = 49° 57' N. 
Z" = 37° CN. 



L d = 12° 57' S. 
= 777 miles. 



To find L m : 
U = 49° 57' N. 
Iff = 37° 0' N. 



57' 



£ m =43°29' 



To find \ d : 

A' = 5° 14' W. 

A" =25° 6' W. 



A d = 19° 52' W. 
= 1192 miles. 



To find the course : 

K cos L„ 



tanO 



[6] 



\ d ==1192, log A d = 3.07628 
L m = 43° 29', log cos L m = 9.86068 
L d =777, cologid = 7.10958 

tan O = 10.04654 
.-. C=S.48°4'W. 



To find the distance : 

D = L d sec0. [2] 

log L d = 2.89042 
log sec (7=0.17505 



log D = 3.06547 
.-. D = 1163 miles. 



26 



NAVIGATION. 



Exercise IV. 




u. 


L", 


\'. 


\". 


C. 


ft 


1. 


25°35'N. 


27° 28' N. 


60° (yw. 


54° 55' W. 


Required. 


Required. 


2. 


32° 30' N. 


34° 10' N. 


25°24'W. 


29° 8'W. 


Required. 


Required. 


3. 


39° 30' S. 


41° O'S. 


74° 2V E. 


70° 12' E. 


Required. 


Required. 


4. 


46° 24' S. 


Required. 


178° 28' E. 


Required. 


S.E.fE. 


278.0 


5. 


20° 29' N. 


Required. 


179°10 / W. 


Required. 


W.byS.JS. 


333.0 


6. 


0° 5& N. 


Required. 


29°50'W. 


Required. 


S. 47° E. 


168.0 


7. 


42° 25* K 


Required. 


66° 14'W. 


Required. 


S.E. by E. 


25.0 


8. 


42° 8'N. 


Required. 


65° 48' W. 


Required. 


E.fS. 


126.0 


9. 


41° 5# N. 


Required. 


62°47'W. 


Required. 


E.|S. 


161.0 


10. 


41° 38' N. 


41° 2& N. 


59° 16'W. 


Required. 


E. by S. 


Required. 


11. 


41° 19* N. 


41°11'N. 


57° 47' W. 


Required. 


Required. 


167.0 


12. 


46° 28' N. 


45°17'N. 


22° 18'W. 


19°39'W. 


Required. 


Required. 


13. 


25° 30' S. 


28° 1& S. 


2°15'E. 


11°17'E. 


Required. 


Required. 


14. 


33°40'N. 


30° 49> N. 


13° 20' E. 


17° 5& E. 


Required. 


Required. 


15. 


19°30'S. 


17°24'S. 


0° 10' E. 


1° 28'W. 


Required. 


Required. 



16. A ship sails from Boston light-house, in latitude 49° 20' 
N., longitude 71° 4' W., on a N.N.E. course, 184 miles. Find 
the latitude and longitude in. 

17. A ship sails from Cape May, in latitude 38° 56' N., lon- 
gitude 74°57'W, on a S.S.E. course, 240 miles. Find the 
latitude and longitude in. 

18. A ship sails from Cape Cod light, in latitude 42° 2' N., 
longitude 70° 3' W., on an E. by N. compass course, 170 miles ; 
wind S.E. by S., leeway \ point, deviation 17f° E., variation 
1H° W. Find the latitude and longitude in. 

19. A ship sails from Cape Cod light on a S.S.E. compass 
course, 140 miles; deviation 5j° E., variation 11?° W. Find 
the latitude and longitude in. 

20. A ship sails from latitude 55° 1» N, longitude 1°25' W. 
on a S.W. compass course, 101 miles; wind W.N.W., leeway 
H points, deviation 6° W., variation 24° 56' W. Find the 
latitude and longitude in. 



mercator's sailing. 27 

21. A ship sails from the Bermudas, in latitude 32° 18' N., 
longitude 64°50'W., on aW.S.W. compass course, 190 miles; 
deviation 1 point W., variation 1 point W. Find the latitude 
and longitude in. 

22. A ship sails from the Bermudas on a W.N.W. compass 
course, 90 miles ; wind S.W., leeway 1 point, deviation 1 point 
E., variation 1 point W. Find latitude and longitude in. 

23. A navigator wishes to sail on a rhumb from the Ber- 
mudas to Cape Fear, in latitude 33° 52' N, longitude 78° W. ; 
variation 10° W., deviation 7° W. Find the compass course 
and distance. 

24. A ship from latitude 36° 32' N. sails between south and 
west until she has made 480 miles of departure, and 9° 22' of 
difference of longitude. Kequired the latitude in, the course 
steered, and the distance run. [Take £„» = £(_£'+-£") + 13'.] 



§ 16. Mercator's Sailing. 
Mercators Chart. 

Mercator's chart represents the surface of the earth as a 
rectangle. The meridians are represented by straight lines 
perpendicular to a straight line representing the equator, and 
the parallels are represented by straight lines parallel to the 
line representing the equator. On this chart, any two meri- 
dians intercept equal distances on all the parallels; but on 
the earth, the intercept on any parallel by two meridians is 
less than the intercept on the equator. 

Hence, at no place except the equator does this chart cor- 
rectly represent the intercepts on the parallels, and the error 
increases as we approach the poles. 

In order to compensate for this error, the distance of each 
parallel from the equator is proportionally increased. 

These augmented latitudes are called Meridional Parts, and 
are given for every latitude in Table XI, 



28 



NAVIGATION. 



..,..,,, , . 


1 1 i 1 


1 1 1 ! ■ 
























- 














- 


\ 


/ j 










- 


% 


fC 




6 






- 




m/ 




















- 


c/ 












- 


A 


























) 1 1 1 1 ... 


> i i i 


1 1 1 1 1 


i i i i 'i 


^ 



55 



45 



25 20 15 10 



5 West Long. 



Fig. 14. 



mercator's sailing. 29 

On account of the spheroidal form of the earth, the develop- 
ment of the formula for the computation of this table is too 
difficult for presentation in an elementary work. 

The meridional difference of latitude is the expanded arc of 
the meridian between two parallels. It may be found by tak- 
ing the difference or sum of the meridional parts correspond- 
ing to the two latitudes, according as the latitudes are of the 
same name or of contrary names. 

If the earth be regarded as a sphere, 

\ d = p sec L. [4] 

But on Mercator's chart, p = X d ; that is, p is too great in 
the ratio of the secant of the latitude. Hence, the intercepts 
of the parallels on the meridians must be increased in the same 
ratio. The augmented latitude for n degrees of latitude would 
be 

1' (sec 1' + sec 2' + sec 3'+ -f seen'). 

The construction and use of Mercator's chart will be under- 
stood by referring to Fig. 14, which represents a chart extend- 
ing from 30° to 60° north latitude, and from 0° to 25° west 
longitude. 

The horizontal line representing the parallel of 30° is first 
drawn, and divided into equal parts. Through the points of 
division the meridians are drawn perpendicular to the paral- 
lel. In this case each division of the parallel represents 
5° = 300'. 

From the Table of Meridional Parts, 

Meridional parts corresponding to a latitude of 30° = 1876.9' 
Meridional parts corresponding to a latitude of 35° = 2231.1' 

Meridional difference of latitude for 5° = 354.2' 

Hence, the parallel of 35° is drawn at a distance of 354.2' 
above the parallel of 30°. 

Meridional parts corresponding to a latitude of 35° = 2231.1' 
Meridional parts corresponding to a latitude of 40° = 2607.9' 

Meridional difference of latitude for 5° = 376,8' 



30 



NAVIGATION. 



Hence, the parallel of 40° is drawn at a distance of 376.8' 
above the parallel of 35°, etc. 

Let it be required to find from the chart the course and 
distance by rhumb-line from A, in latitude 35° N., longitude 
20° W., to B, in latitude 50° N., longitude 5° W. 

A is at the intersection of the meridian of 20° and the par- 
allel of 35° ; B is at the intersection of the meridian of 5° and 
the parallel of 50°. Draw AB. AB represents the rhumb- 
line from A to B, for it makes the same angle ((?) with all of 
the meridians. 

The course C may be found by making one edge of a pair 
of parallel rulers coincide with AB, and extending until the 
other edge passes through the centre of the compass-rose above. 
The rhumb-line of the compass-rose which most nearly coin- 
cides with the edge of the ruler will show the course. 

The distance may be found by carrying m, the middle point 
of AB, over to the side, and measuring up and down a distance 
equal to one-half of AB. The number of degrees between the 
extreme points reduced to miles will be the distance required. 
It is evident that Mercator's chart shows relative positions 
accurately, but not relative distances. 

Let CA (Fig. 15) represent the ship's track. CB will rep- 
resent the true difference of latitude, and AB 
the departure. Take CE equal to the merid- 
ional difference of latitude corresponding to 
CB, and through E draw ED parallel to BA 
to meet CA produced in D. Then CD will 
represent the ship's track on Mercator's chart, 
and ED will represent the difference of lon- 
gitude. 

CE= Mer. L d 

= meridional difference of latitude. 
CB = L d . BA=p. ED = X d . 




Fig. 15. 



In the triangle CED, 
tan C— 



ED 

EC' 



MERCATOB S SAILING. 



31 



or 



tanO = 



Also, 



Also, 



Mer.L; 
/. X d = Mer. L d X tanO. 

tanO=£. 
L> d 

D = K, sec 0. 



[8] 

[3] 

[9] 
[2] 



When to use Middle Latitude and Mercator's Sailings. 

Either Middle Latitude or Mercator's Sailing may be used 
when the distance is small (such as an ordinary day's run); 
but when the distance is great, Mercator's Sailing should be 
employed, if the course be less than about 45° ; and Middle 
Latitude Sailing should be employed if the course exceed 
about 45°. 

This distinction is of special importance in high latitudes. 
Example 1. A ship sails from latitude 40° 45' N., longitude 
74° W, E.N.E., 200 miles. Find latitude and longitude in. 
Solution I. By computation. 

To find longitude in : 
L< =40°45 / N., Mer. Parts = 2666.8 
£" = 42° 2'N., Mer. Parts = 2769.0 

Mer. L d = 102.2 

X d = Mer.L d xtan0. [8] 

Mer.Z d =102.2, log Mer. L d = 2.00945 
0= 67° 30 7 , log tan O = 10.38278 

log Ad = 2.39223 

.-. \ d = 246.7 miles 
= 4° 7'E. 
a' = 74° 0' W. 

\ff= 69° 53' W. 

Solution II. By inspection. E.N.E. = 6 points. Corresponding to 
a course of 6 points and a distance of 200, the true difference of latitude 
is found in Table VIII. to be 76.5; whence, the latitude in and the meri- 
dional difference of latitude are found as in Solution I. Corresponding to 





To find latitude in : 








L,= 


= DcosO. 


[2] 


D 


= 200 




logD - 


2.30103 


C 


= 67° 


30 7 , 


log cos Q= 
log L d = 


9.58284 




1.88387 




• 


:L a 


= 76.5 miles 
= 1°17'N. 








L> 
L" 


= 40° 45' N. 






= 42° 2'N. 





32 



NAVIGATION. 



the course 6 points, and the meridional difference of latitude 102 as a 
difference of latitude, the same table gives the departure 246.7, which is 
equal to the difference of longitude required ; whence, the longitude in 
may be found. 

Example 2. Required the course and distance from Hali- 
fax, in latitude 44° 40' N. and longitude 63° 35' W., to Ireland 
island, in latitude 32° 19' N. and longitude 64° 49' W. 



L' = 44° 4(y N. 
L" = 32° 19' N. 



Mer. Parts = 2985.6 
Mer. Parts = 2038.6 



\' =63°35 / W. 
a" = 64° 49' W. 



L d = 12° 21' S. 
= 741 miles. 

To find the course 



Mer.Z,* = 947.0 



A d = 1°14'W. 
= 74 miles. 



Mer. 



= 74, 
i d = 947 



tanO= A * , 
Mer. L d 

logA d 



To find the distance : 

L d = 741, 
C =4° 28', 



[8] 



= 1.86923 
colog Mer. L d = 7.02365 

log tan C 



= 8.89288 
C=S. 4° 28' W. 



D = L d sec 0. 



log L d = 2.86982 
log sec (7=0.00132 

log D = 2.87114 
.-. D = 743.3 miles. 



[2] 



Exercise V. 








L'. 


L". 


K\ 


A.". 


C. 


D. 


1. 


38°14'N. 


39° 51' N. 


2° 7'E. 


4° 18' E. 


Required. 


Required. 


2. 


49° 53' N. 


48° 28' N. 


6°19'W. 


5° 3'W. 


Required. 


Required. 


3. 


64°30 / K 


60° 40' N. 


4°20'W. 


0°1(/E. 


Required. 


Required. 


4. 


54° 54^ S. 


34° 22' S. 


60°28'W. 


18° 24 W. 


Required. 


Required. 


5. 


17° CN. 


20° O'N. 


180° 0'E. 


177° 0'E. 


Required. 


Required. 


6. 


45° 15> N. 


Required. 


35°26'W. 


Required. 


N. 49° E. 


175 


7. 


55° l'N. 


Required. 


1°25'E. 


Required. 


N. 10° E. 


246 


8. 


50° 48' N. 


Required. 


9°i(yw. 


Required. 


S. 41° W. 


275 


9. 


37° O'N. 


51° 18' N. 


48°2(yW. 


Required. 


Required. 


1027 


10. 


51° 15' N. 


37° 5'N. 


9°50'W. 


Required. 


S.W.byS. 


Required. 



TRAVERSE SAILING. 33 

11. Required the course and distance from Toulon to Val- 
encia, by Mercator's sailing : 

Tonlon{ x= &0 ^ R Valencmj^ ^^ 

12. Required the compass course and distance from Cape 

East, New Zealand, to San Francisco; variation 14° 20' E., 

and deviation 5° 40' E. : 

r L= 37°40'S. a - „ . ( L= 37°48'N. 

Cape Easts . ,►„■*„« «,t, San Francisco < „ „ * 

F U = 178°36'E. I X = 122°24'W. 

13. Required the course and distance from Cape Lopatka 
to Callao : 

n T ., (L= 51° 2'N. n ,, J £=12° 4'S. 
Cape Lopatka{ x ^^ Calla ° { A = 77° 14' W. 

14. A ship from latitude 20°40'N. sails N.E. by N. until 
she is in latitude 27° 16' N. Required the distance and differ- 
ence of longitude. 

15. A ship from Cape Clear, in latitude 51° 26' N. and lon- 
gitude 9° 29' W., sails S.W. by S. until the distance run is 1022 
miles. Find the latitude and longitude in by Mercator's and 
Middle Latitude Sailings. Which method is preferable ? 

§ 17. Traverse Sailing. 

When a ship reaches her destination by sailing on different 
courses, the difference of latitude may be found by computing 
the change in latitude for each course separately, as in Plane 
Sailing, and taking the algebraic sum ; or, more conveniently 
(but less accurately), by writing the north latitudes in one 
column and the south latitudes in another, finding the sum of 
each column and taking the arithmetical difference, giving it 
the name (N. or S.) of the greater sum. In like manner, the 
total departure may be found approximately. 

Evidently the Traverse Table may be used, and the work 
arranged as in Rectangular Surveying. 

When the difference of latitude and departure have been 
found, the difference of longitude may be determined by Mid- 
dle Latitude or Mercator's Sailings. 



34 



NAVIGATION. 



Example 1. A ship in latitude 40° N. and longitude 67° 4' 
W., sails N.W. 60 miles, then N. by W. 52 miles, then W.S.W. 
83 miles. Find the latitude and longitude in. 



c. 


D. 


N. 


5. 


E. 


w. 


N.W. 
N. by W. 
W.S.W. 


4 pts. 
1 " 
6 " 


60 
52 
83 


42.4 
51.0 


31.8 




42.4 
10.1 
76.7 


Hence, i d = 61.6 miles N. 
= 1° 2' N. 


93.4 
31.8 


31.8 





129.2 
0. 


P 


= 129.2 


W. 


61.6 


129.2 





IS. 


24. 


V. 




V = 40° 7 N. 


2/ = 40° <y N. 


x' = 67° 4' W: 




i d = 1° 2f ]S T . 


2/' =41° 2 / N. 


\ d = 2° 51' W. 




I/' = 41° 2f N. 


2)S1° 2' 
24 = 40° 31' 


A" = 69° 55' 



90° 



co L m = 49° 29 / 

The difference of longitude may be found by computation, using form- 
ula [51, or by inspection, as in \ 15, Ex. 2. 49° is found at the bottom 
of page 118, Table IX., and opposite the departure 129.1 is distance 171, 
which is the difference of longitude required. 171' = 2° 51'. 

To Plot the Courses. 
Let (7 (Fig. 16) be the 
centre of a circle repre- 
senting the card of the 
compass. Draw CA, N.W. 
from C, equal to 60, to any 
convenient scale. Make 
an edge of a pair of par- 
allel rulers coincide with 
the N. by W. rhumb of 
the compass diagram, ex- 
tend until the other edge embraces A, then draw AB equal 
to 52. Make an edge of the ruler coincide with the W. S.W. 




Fig. 16, 



TEAVERSE SAILING. 



35 



rhumb of the compass diagram, and extend until the other 
edge embraces B, and draw BD equal to 83. D will represent 
the position of the vessel. 

Upon CN produced let fall the perpendicular DF. Then 
CFwill represent the total difference of latitude, CD the dis- 
tance made good, and DCF the course made good. 

The departure may be regarded as measured on the parallel 
HB' + L"). 

Example 2. The U.S. Steamer Swatara left latitude 37° 25' 
N. and longitude 3° 14' E., at noon, Feb. 25, 1880, and during 
the next 24 hours sailed on the (corrected) courses in the fol- 
lowing tablet. Find the latitude and longitude in by account, 
at noon, Feb. 26. 



c. 


D. 


N. 


S. 


E. 


w. 


N.E. by E. f E. 


5| pts. 


2.0 


0.9 




1.8 




E. by N. 


7 " 


28.0 


5.5 




27.5 




E.JS. 


7J " 


3.0 




0.3 


3.0 




E.S.E. J E. 


6* " 


2.8 




0.8 


2.7 




S.S.E.JE. 


2£ » 


3.4 




3.0 


1.6 




N. by W. \ W. 


U." 


4.2 


4.1 






1.0 


N.JE. 


i " 


4.4 


4.4 




0.2 




N. by E. 1 E. 


H " 


10.6 


10.3 




2.6 




N.JE. 


i " 


11.4 


11.4 




1.1 




N.JW. 


i " 


18.6 


18.5 






1.8 


N. by W. 


l «« 


34.8 


34.1 






6.8 


S. by E. } E. 


i* " 


5.0 




4.8 


1.5 








89.2 


8.9 


42.0 


9.6 


Hence, L d = 80.3 = 1° 
and p == 32.4 E. 


20 / N., 


8.9 




9.6 
















80.3 




32.4 





L". 
U = 37°25'N. 
L d = l°2(yN. 

2/' = 38° 45' N. 



L> = 37° 25' N. 
L" = 38° 45' N. 

2)76° 10' 

Z TO = 38° 5' 

coi w = 51°55' 



A' = 3° 14' E. 
\ d = 0° 41' E. 

A" = 3° 55' E. 



36 



NAVIGATION. 



Exercise VI. 
Required the difference of latitude and the departure 



1. 



c. 


D. 


S.S.W. 
S.W. by S. 
N.E. 


48 
36 
24 



2. 



c. 


D. 


S.JE. 
S.W.JS. 
S.S.W. i W. 


18 
37 
56 



3. 



c. 


ft 


S.S.W. i W. 

s.s.w. -i-w. 

S. by W. J W. 


43 
39 

27 





4. 




c. 


D. 


N. 


25° W. 


16.4 


N. 


8°E. 


7.8 


N. 


19° E. 


13.7 


N. 


76° E. 


39.6 



5. 




c. 


ft 


W.N.W.JW. 
N.N.E.fE. 
N. by E. f E. 
S.S.W. \ W. 


21 

9 

9 

30 



G, 


D 


S. 


83° W. 


23.0 


s. 


4S°E. 


25.2 


N. 


48° W. 


27.1 


N. 


36° W. 


21.0 



7. 



c. 


ft 


S. 17° E. 


48 


S. 45° W. 


19 


N. 36° W. 


18 


N. 41° W. 


50 


E. 


36 



8. 




c. 


ft 


N.N.E. 


31 


E.N.E. 


35 


E. by S. 


36 


S.S.E. 


51 


S. by E. 


60 



a 


D. 


S. 44° E. 


69 


S. 85° E. 


68 


S. 27° E. 


25 


N. 37° W. 


5 


N. 20° W. 


13 



§ 18. The Bay's Work. 

When a ship is starting out upon a voyage, just before she 
passes out of sight of land, the proper officer notes the bearing, 
and estimates the distance from the ship, of some object on 
shore whose latitude and longitude are known. This opera- 
tion is called Taking the Departure. After this has been done, 
a record is kept on the log-board of the courses and distances 
run. From the log-board the record is transferred to the log- 
book, in which it is arranged as in the following tablets. The 



the day's work. 37 



column headed "H" contains the hours from noon to noon. 
The column headed "K" contains the knots per hour the 
ship has sailed ; and the next column contains the tenths of 
knots. The remainder of the tablet does not require explana- 
tion. 

From the record of the log-book the position of the ship is 
computed by Traverse Sailing each day at noon, or at any 
other hour when the position is desired, in the manner 
described in the following examples. 

In computing the day's work, the first course of the traverse 
is found by reversing the departure course. For, if, as in 
Example 1, the bearing of the object from the ship be N.E. \ E. 
15 miles, then it follows that the bearing of the ship from the 
object on shore is S.W. \ W. 15 miles. 

This method of finding the place of a ship is subject to many 
errors. It is impossible to determine the course accurately: 
the deviation is almost constantly changing ; the effect of tides 
and currents can be estimated only approximately ; the log 
gives the distance roughly ; the speed varies during the hour ; 
etc., etc. 

Hence, whenever it is possible to do so, the results obtained 
by dead-reckoning should be verified by celestial observation. 
(See Chapter III.) The results obtained by the latter method 
should be given the preference. 

A number of the following exercises are taken from recent 
examination papers of the British Marine Board ; others are 
actual transcripts from the log-books of vessels. 

The log-book used in the U. S. Navy contains, in addition 
to the data of the following tablets, columns in which are re- 
corded the force of the wind, the height of the barometer, the 
temperature of the air by dry and wet bulb thermometers, 
the surface temperature of the water, the state of the weather, 
the forms of clouds, the proportion of clear sky, the state 
of the sea, a record of the sail the vessel is under at the end of 
the watch ; also, a record of the miscellaneous events of the 
day. 



38 



NAVIGATION. 



Example 1. 



H. 


Courses, 


K, 


-h 


W/Ws, 


Leeway, 


Deviation, 


Remarks, etc. 


1 
2 


w.s.w. 


10 
11 


8 
4 


N.W. 


|pt. 


11° w. 


The departure \^s 1 
taken as follows : A 










point of land in lati- 


3 




11 


4 








tude 37° 3' N., longi- 


4 




11 


4 








tude 9° O' W., bear- 


5 


N.W. \ N. 


12 


2 


W.S.W. 


Ipt. 


17° W. 


ing by compass N.E. 
34 B., dist. 15 miles. 


6 




12 


3 








Ship's head W.S.W. 


7 




12 


3 








Deviation as per log. 


8 




12 


2 










9 


N.N.W. 


9 


6 


W. 


f pt. 


11° W. 




10 




9 


5 










11 




9 


5 










12 




9 


4 








Variation 22° 30' W. 


1 

2 
3 


N.W. by W. 


7 
7 
7 


8 
6 


S.W. by W. 


1| pts. 


20° W. 






4 








A current set the 


4 




8 


2 








ship S. W. by W. 


5 


S.W. \ S. 


9 


3 


S.S.E. 


lpt. 


6° W. 


% W. (correct mag- 
netic) 8 miles, from 


6 




8 


7 








the time the depar- 


7 




9 


3 








ture was taken to the 


8 




8 


7 








end of the day. 


9 


w.j a 


10 


3 


S.byW. 


Jpt 


13° W. 




10 




10 


2 










11 




10 


2 










12 




10 


3 











The departure course. 

The opposite point to N.E. \ E. 
is S.W. \ W. The ship's head be- 
ing W. S.W., the deviation is the 
same as for the first course, viz., 
11° W. 



S.W.JW. =4ipts.R.ofS. 

or 47° 49' R. of S. 

Dev.ll 0'W. 
Var. 22° 30' W. 
True course 



33° 30' L. 



= 14°19 / R. ofS. 

Hence, the first course and dis- 
tance of the traverse are 

S. 14° W. 15 miles. 



First course. 

W.S.W. 
Leeway 

or 
Dev.ll 0'W. 
Var. 22° 30' W. 
True course 



= 6 pts. R. ofS. 
- } " L. 

5£ pts. R. of S. 
61°53'R. of S. 

= 33°30 / L. 

= 28°23'R. of S. 



The distance (45 miles) for this 
course is found by adding together 
the distances sailed each hour up 
to 5 o'clock, when the course was 
changed. 

Hence, the second course and 
distance of the traverse are 
S. 28° W. 45 miles. 



THE DAY S WOEK. 



39 



Second course. 
N.W. £ N. 
Leeway 



or 
Dev. 17° O'W. 
Var. 22° 30' W. 
True course 



= 3Jpts. L.ofN. 
= i " B. 

3|: pts. L. ofN. 
36° 34' L. of N. 

= 39° 30' L. 

= 76° 4'L.ofN. 



The distance (49 mi.) is obtained 
by adding together the hourly dis- 
tances from 5 to 9 o'clock. 

Hence, the third course and dis- 
tance of the traverse are 

N. 76° W. 49 miles. 



Third course. 

N.N.W. 
Leeway 



or 
Dev. 11° O'W. 
Var. 22° 30' W. 
True course 



= 2 pts.L.ofN. 
- f " R. 

1£ pts. L.ofN. 

14° 4'L.ofN. 

= 33° 30' L. 

= 47°34'L. ofN. 



The distance (38 mi.) is obtained 
by adding together the hourly dis- 
tances from 9 to 1 o'clock. 

Hence, the fourth course and 
distance of the traverse are 
N. 48° W. 38 miles. 



Fourth course. 
N.W. by W. 
Leeway 



or 
Dev. 20° O'W. 
Var. 22° 30' W. 
True course 



5 pts.L.ofN. 
\\ " R. 

3fpts. L.ofN. 
42° 11' L. ofN. 

42° 30' L. 
84° 41' L.ofN. 



Hence, the fifth course and dis- 
tance of the traverse are 

N. 85° W. 31 miles. 



Fifth course. 
S.W. I S. 
Leeway 



or 

Dev. 6° O'W. 
Var. 22° 30' W. 
True course 



= 3£pts.R. ofS. 
= 1 " R. 

4 J pts. R. ofS. 
50° 38' R. of S. 

= 28°30'L. 



= 22° 8'R. ofS. 
Hence, the sixth course and dis- 
tance of the traverse are 
S. 22° W. 36 miles. 



Sixth course. 

W. }S. 

Leeway 

or 
Dev. 13° O'W 
Var. 22° 30' W 
True course 

Hence 



= 7£pts.R.ofS. 

- j " R. 

8 pts. R. ofS. 
90° 0'R. ofS. 

- 35° 30' L. 



= 54°30'R.ofS. 
the seventh course and 
distance of the traverse are 
S. 55° W. 41 miles. 



Current course. 
S.W. by W. f W. 

or 
Variation 



5|pts. R. ofS. 
64° 41 'R. of S. 
22° 30 7 L. 



True course = 42° 11' R. of S. 

Hence, the eighth course and 
distance of the traverse are 
S. 42° W. 8 miles. 



40 



NAVIGATION. 



The To-averse. 





Courses. 


Dlst. 


«. 


s. 


E. 


W. 


1 


S. 14° W. 


15 




14.6 




3.6 


2 


S. 28° W. 


45 




39.7 




21.1 


3 


N. 76° W. 


49 


11.9 






47.5 


4 


N. 48° W. 


38 


25.4 






28.2 


5 


N. 85° W. 


31 


2.7 






30.9 


6 


S. 22° W. 


36 




33.4 




13.5 


7 


S. 55° W. 


41 




23.5 




33.6 


8 


S. 42° W. 


8 




5.9 




5.4 


! 2,^=77.1 miles 


40.0 


117.1 




183.8 


= 1° 17' S., 




40.0 






r* 


= 183.8 W 




77.1 



To find the latitude in 



L' =37° 


3' N. 


L d = 1° 


17' S. 


L" = 35° 


46' N. 


Also, i w =36° 


25'. 


To find the direct 


course : 


tanO - 


_ P 



logp = 2.26435 
cologid = 8.11295 

log tan C= 10.37730 

.-. C = S.67°W 



[3] 



[1] 



To find the direct distance : 

sinu 
logp = 2.26435 

cologsin 0=0.03517 

log D = 2.29952 
/. D = 199 miles. 



To find the difference of longitude 
A d = psecL m . *[4 

logp = 2.26435 

log sec L m = 0.09435 

logA d =2.35870 

.-. \ d = 228' = 3° 48' W. 



To find the longitude in : 

x' = 9° (y w. 

\ d = 3°48'W. 
A" = 12° 48' W. 



THE DAY S WORK. 



41 



Example 2. 



H. 


Courses. 


K. 


A 


Winds. 


Leeway. 


Deviation. 


Remarks. 


1 

2 


S.W. 


6 

6 



3 


W.N.W. 


lpt. 





Latitude from 28° 
30' S. 
Longitude from 


3 




6 


4 








10° 15' E. 


4 




6 













5 




5 


3 










6 

7 
8 


S. by W. 


6 
5 
5 



1 

4 


W.byS. 


lpt. 
















9 




5 


2 










10 




5 


3 








Variation 1% pts.W. 


11 




5 













12 


S.S.W. 


5 


2 


W. 


1 pt. 






1 




5 


2 










?, 




5 













3 




4 


6 








A current set S. 










(correct magnetic) 


4 




b 











2% miles per hour, 


B 




5 











from noon to noon. 


6 
7 
8 


S.E. by S. 


5 
5 
5 


2 
4 
5 


S.W. by S. 


lpt. 
















9 




6 













10 




6 













11 




5 


4 










12 




5 


5 











First course. 




Third course. 




S.W. 
Leeway 


= 4 pts. R. of S. 
= 1 " L. 


S. S.W. 
Leeway 

Variation 


= 2 pts. R. of S. 
= 1 " L. 


Variation 


3 pts. R. of S. 
-lj " L. 


1 pt. R. of S. 
= 11 " L. 


True course 


= 1 J pts. R. of S. 
-S. ljpts. W. 


True course 


= £ pt. L. of S. 
= S. J pt. E. 


Second course. 




Fourth course. 




S. by W. 
Leeway 


= 1 pt. R. of S. 
= 1 " L. 


S.E. by S. 
Leeway 

Variation 


= 3 pts. L. of S. 
= 1 " L. 


Variation 


pts. R. of S. 
= 1* " L. 


4 pts. L. of S. 
= 1* " L. 


True course 


= 1£ " L. ofS. 
- S. l\ pts. E. 


True course 


= 5£ pts. L. of S. 
= S. 5£ pts. E. 






42 



NAVIGATION. 



The current course. 
S. =0 pts. L. of S. 

Variation = 1£ " L. 



True course = 1£ pts. L. of S. 
= S. 1J pts. E. 



The Traverse. 





Courses. 


Dist. 


N. 


S. 


E. 


W. 


1 


S. 1| pts. W. 


30 




28.7 




8.7 


2 


S. 1£ pts. E. 


32 




30.6 


9.3 




3 


S. |pt. E. 


30 




29.9 


2.9 




4 


S. 5| pts. E. 


39 




18.4 


34.4 




5 


S. li pts. E. 


60 




57.4 


17.4 




L d = 165 miles = 2° 45' S. 




165.0 


64.0 


8.7 


£' 


= 28° 3(X S. 






8.7 




Z" 


= 31° 15' S. 


55.3 



1/ = 28° 30 7 S. 
£" = 31° 15' S. 



Mer. Parts 
Mer. Parts 



Mer. L d 



1774.3 
1963.6 

189.3 



To find the difference of longitude : 

. Mer. L d x p 



[9] 



Mer. X d = 189.3, 
p = 55.3, 
£ d = 165.0, 



Arf 



log = 2.27715 
log = 1.74273 
colog = 7.78252 

log \ d = 1.80240 
63.4 = 1° 3' E. 



\l = 


10° 


15' 


E. 


A d = 


1° 


3' 


E. 


\" = 


11° 


IS' 


E. 



The distance made good may be found by [2], and the course made 
good by [3] or [8]. 

EXEECISE VII. 

Find the latitude and longitude in from the following log- 
book records: 



THE DAY S WORK. 



43 



1. 



H. 


Courses. 


K, 


r. 


Winds. 


/.eeway, 


Deviation. 


Remarks. 


1 

2 


N.N.E. 


6 
6 


3 
2 


w. 








Latitude from 46° 
28' N. 

Longitude from 
22° 18' W. 


3 




6 


5 










4 




6 


4 










5 




6 













6 


E.N.E. 


6 


1 


N.W. 








7 




6 


6 










8 




5 


8 










9 




5 


6 










10 




5 


4 










11 




5 


5 










12 


E. by S. 


5 


3 


N. 






Variation 0. 


1 




5 


9 










2 




6 


2 










3 




6 













4 




6 


3 










5 




6 


4 










6 

7 


S.S.E. 


7 
6 




8 


N. 






The tide set S. by 
E., 2}4 mi- per hour, 
during the 24 hours. 


8 




7 


3 










9 




7 


5 










10 




7 


1 










11 




7 


9 










12 




7 


3 











44 



NAVIGATION. 



Courses. 



Leeway. 



Deviation. 



Remarks. 



9 

10 

11 

12 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 



S. by W. 



S.W. by S. 



S.W.byW. 



W. 



Latitude from 33° 
40' N. 

Longitude from 
16° 20' "W. 



W. by N. 



lpt. 



Variation 12° 20' W. 



N.W. 



A current set W 
S.W. (correct mag 
netic), 1J4 miles per 
hour from noon to 
noon. 



THE DAY S WORK. 



45 



3. 



H. 


Courses. 


ft; 


r. 


Winds. 


Leeway. 


Deviation. 


Remarks. 


1 


N. by E. 


6 


7 


E. by N. 


1 pt. 





Latitude from 19° 
30' S. 


2 




6 


2 








Longitude from 0° 
10' E. 


3 




6 


4 










4 




6 


3 










5 




6 


1 










6 




6 













7 


N. 


5 


8 


E.N.E. 


lpt. 






8 




5 


4 










9 




5 













10 




5 


3 










11 




5 


6 










12 




5 


9 








Variation 13° 30' W. 


1 




5 


7 










2 


N.N.W. 


6 


4 


N.E. 


lpt. 






3 




6 


8 










4 




7 













5 




7 


3 










6 

7 
8 




7 
7 
7 


6 
5 









The current set W. 
N. W. (correct mag- 
netic), %< mile per 
hour, from noon to 
noon. 


9 




7 


2 










10 




7 


4 










11 




6 


3 










12 




6 














46 



NAVIGATION. 



4. 



H. 



Courses. 



Winds. 



Leeway. 



Deviation. 



Remarks. 



S.E. by E. 



S.E. 



E. by N. 



E.N.E. 



S.S.E. 



S.E. by S. 



E.N.E. 



|pt 



S.E. by S. lpt. 



S.E. 



E. by N. 



Hpto 



2pts. 



lipts. 



11° E. 



9°E. 



17° E. 



15° E. 



8°E. 



A point of land, 
latitude i~v 31' N., 
longitude 52° 33 "W., 
bearing by compass 
"W.S.W., distant 18 
miles. Ship's head 
S.E. by E. Dela- 
tion as per log. 



Variation 25" AV. 



A current set the 
ship (correct mag- 
netic) S. by E. 12 
miles from the time 
the departure was 
taken to the end of 
the day. 



THE DAY S WOEK. 



47 



H. 


Courses. 


K. 


T. 


Winds. 


Leeway. 


Deviation. 


Remarks. 

4 


1 
2 
3 
4 


S.S.E. 


5 
5 
5 
4 


6 
6 


8 


E. 


2£pts. 


3°E. 


A point of land in 
latitude 62° N., lon- 
gitude 150° E., bear- 
ing by compass W. 
by S. X S., distant 
17 mi. Ship's head 
S.S.E.; deviation as 
per log. 


5 


S.S.W.JW. 


4 


7 


w. 


2| pts. 


4°W. 




6 




4 


8 










7 




5 


2 










8 




5 


3 










9 


w.s.w. 


5 





s. 


2£ pts. 


9°W. 




10 




6 













11 




6 


5 










12 




6 


5 










1 


W.*N. 


6 


6 


K by E. 





11° W. 


Variation 31° E. 


2 




7 













3 




6 


4 










4 




6 













5 


E. 


4 


6 


S.S.E. 


2£ pts. 


10° E. 




6 

7 
8 




5 
4 
4 



8 
6 








A current set the 










ship (correct mag- 
netic) N.N.E. 21 mi. 
from the time the 
departure was taken 
to the end of the day. 


9 


E.S.E. 


4 





S. by W. 





9°E. 




10 




4 


5 










11 




4 


5 










12 




5 














48 



NAVIGATION. 



H. 


Courses. 


A-. 


T. 


H7aifa 


Leeway. 


Deviation, 


Remarks. 


1 

2 
3 
4 


s.w. i w. 


14 
14 
14 
14 


5 
2 
6 

7 


S.E. 





6° W. 


A point, latitude 
50° 12' S., longitude 
179° 40' W., bearing 
by compass N.%W., 
distant 19 miles. 

Ship's head S.W. 
%W. Deviation as 
per log. 


5 


N.fE. 


4 





E.N.E. 


3^ pts. 


3° E. 




6 




3 


6 










7 




3 


6 










8 




3 


8 










9 


S.byE.JE. 


2 


4 


S.W. \ W. 


2£pts. 


6°E. 




10 




2 


3 










11 




2 


3 










12 




2 













1 


W. by S. 


12 


2 


S.byW. 


ipt. 


14° W. 


Variation 14° E. 


2 




12 


4 










3 




12 


6 










4 




12 


8 










5 
6 
7 
8 


E.N'.E. 


3 
2 
3 
3 



3 

4 
3 


S.E. 


2£ pts. 


19° E. 


A current set the 
ship (correct mag- 
netic) S.W. K W. 42 
miles from the time 
the departure was 
taken to the end of 
the day. 


9 


S.S.W.£W. 


5 


6 


S.E. 


If pts. 


4° W. 




10 




5 


7 










11 




5 


3 










12 




5 


4 











THE DAY S WORK. 



49 



H. 


Courses, 


K. 


7". 


Winds, 


Leeway, 


Deviation, 


Remarks, 


1 

2 














Left )Z36°42'N. 
Malaga ^ A 4 o 25 > w . 
at 8 p.m. J 


3 
















4 
















5 
















6 
















7 
















8 


S.E. 


1 


5 







I pt. E. 




9 




6 


4 










10 




6 


6 










11 




6 


8 










12 




6 


5 










1 


E.S.E.£E. 


7 


2 










Variation 1% pts. W. 


2 




7 


2 










3 




8 


2 










4 




8 


6 










5 




8 













6 


• 


7 


6 










7 




7 


6 










8 




7 


6 










9 




7 


5 










10 




5 


7 










11 


E. 


4 


2 







£ pt. E. 




12 




4 


5 











50 NAVIGATION. 



§ 19. Great Circle Sailing. 

It may be proven that the shortest distance between two 
places on the surface of the earth is the less of the two arcs of 
a great circle lying between these places. The' rhumb-line is 
the arc of a great circle only when it coincides with the equa- 
tor or a meridian. 

The advantages of great circle sailing are : 

1. The distance is less than by any other route. 

2. The great circle track gives the real direction of the 
desired port, and this frequently determines which tack the 
ship should be put on. 

To sail on a great circle would require a constant change 
of course. This difficulty is practically overcome by sailing 
from point to point of the great circle on rhumb-lines. 

Definition. The vertex of a great circle track is the point 
nearest the pole. Hence, the vertex is the foot of the perpen- 
dicular let fall upon the great circle from the pole. 

The best practical method of finding the courses, distance, 
etc., of the great circle track is by the use of Godfray's Course 
and Distance Diagram, which affords a simple graphic solution. 

The solution by Spherical Trigonometry is given here. 

Let CC (Fig. 17) represent the great circle track between 
C and C, P the pole, and PC and 
PC meridians. 

The angle C = the first course 
from C 

C = the first course from C. 
P = X d z= the difference of longi- 
tude of C and C. 

PC = e 1 = the polar distance of C 
= 90° =F latitude of C 
PC'= c = the polar distance of C = 90° =F latitude of C. 
CC = D = the length of the great circle track. 




GREAT CIRCLE SAILING. 



51 



To find the first course. 

In the triangle CPC (Fig. 17) by Napier's Analogies (Spher. 
Trig. §53): 

tan } (0 + C) = C0S H°- C |) C ot \ \ d , 



cos £ (c + c') 

tanUO-C0 = S | n ^ C 7 C % otU d 
sin \ (c + c') 

o=i(0+o')+^(o-d f ), 

C'=§(0 + 0')-n0-0'). 



[10] 



To find the distance. 
By Gauss's Equations (Spher. Trig. § 53) : 

C o^D = c ° s ^+ c ;) S inU, 
cos I (0 + 0') 

See also Spher. Trig. § 62. 



[11] 



To find the latitude of the vertex. 

Let P V (Fig. 18) be the perpendicular from P upon the 
arc of a great circle CC. V is the p 

vertex of CC. 

In the right triangle CPV, 

sin PV = sine' sin 0, 1 

Latitude of V = 90° -PV. J 

See Spher. Trig. § 47. 



[12] 



To ymc? £Ae longitude of the vertex 

from C 
In the right triangle CPV, 

cos c' = cot (7 cot CPV. 
Hence, cot OP V = cos c' tan 0. 




Fig. 18. 



[13] 



Combining this result with the longitude of C, the longitude 
of T^is found. 



52 NAVIGATION. 






To find the latitude of intermediate points of given longitude. 

Let m (Fig. 18) be a point whose longitude from V is 
assumed. 

In the right triangle PmV, 

cos mP V= tan P Fcot Pm. 



Hence, cot Pm = cos mPV cot PV, 



-PmJ M 



Latitude of m = d= (90° - Pm) 

Points m, etc., are usually assumed on each side of V, dif- 
fering from Fand from each other 5° or 10° in longitude. In 
this manner, the latitude and longitude of as many points as 
we please between C and C may be determined. The course 
and distance from point to point by rhumb-line may then be 
found by Middle Latitude or Mercator's Sailings. 

When either of the angles C or C is greater than 90°, and 
the other is less than 90°, the vertex will be on CO' produced 
through the vertex of the angle which exceeds 90°. (See 
Fig. 20.) 

Note. If 11 and L" represent the latitudes of C and C respectively, 
the first formula of set [10] becomes 

t anKo + co= :°:|g;g;j cot^, 

'provided both places are on the same side of the equator. The student will 
find it a valuable exercise to express all of the preceding formulas of 
this section similarly. 

To plot the Cheat Circle Track on Mercators Chart. 

1. By means of great circle charts. On these charts the 
great circle is represented by a straight line ; hence, the lati- 
tude of each intersection with the meridians is known. The 
corresponding points may be marked on the meridians on 
Mercator's chart and a line traced through these points. This 
curved line will represent the great circle track. 

2. By computing the latitudes corresponding to certain lon- 
gitudes (or vice versa) of the great circle, marking these points 
on Mercator's chart, and tracing the curve through them. 



GREAT CIRCLE SAILING. 



53 



Example 1. Required the initial courses, the distance, and 
the latitude and longitude of the vertex of the great circle 
track joining C and 0". Given: 

Lat.ofC = 54°18'N. 
Lat.of C = 7°12'S. 
Long, of C =142° 38' E. 
Long, of C'= 90° O'W. 

Referring to the triangle 
CPC (Fig. 19), 

c =90°+ 7° 12' = 97° 12', 
c' = 90° - 54° 18' = 35° 42', 
x d = P=(180°-142°38') 
+ (180° -90°) 
= 127° 22'. 

To find the initial courses 




Fig. 19. 



ta n HO + O0 = cos ^ c 7 c V u„] 

cos h (c + c') 

tanKQ-0') = s ? n ^ c 7 c 2 com, 
sin I (c -f o') 

log cos = 9.93420 

colog cos = 0.39843 

log cot = 9.69425 

log tan 



[10] 



30° 45' 
66° 27' 
63° 41' 



10.02688 



Kc + c') 

£(C+C") = 46°46' 
£(<7-C") = 15°25', 

.-. C=62°ll',orN. 62°11'E. 
and C" = 31° 21', or N. 31° 21' W. 

To find the distance : 

008^(0+0') 

i(c + c') = 66° 27' 

|(C+C")= 46° 46' 

^A rf = 63° 41' 

£D = 58° 29', 
.-. Z> = 116°58' 
= 7018 miles 



log sin =9.70867 

colog sin =0.03777 

log cot = 9.69425 

log tan = 9.44069 



[11] 



log cos = 9.60157 

colog cos = 0.16433 

log sin = 9.95248 

log cos = 9.71838 



54 



NAVIGATION. 



To find L of the vertex : 

sin PV = sine' sin 0. [12] 

c' = 35° 42', log sin = 9.76607 
= 62° 11', log sin = 9.94667 

PF=31° 4', log sin = 9.71274 
L of V= 90° - 31° 4' = 58° 56' N. 



To find A of the vertex : 

cot CPV = cose' tan 0. [13] 

log cose 7 = 9.90960 
log tan O = 10.27769 

log cot CPF= 10.18729 
.-. CPV= 33° 1' 
\ of (7-142° 38' E. 



\ofF=175°39'E. 
Example 2. The Cunard steamers in sailing from New York 
to Liverpool keep approximately on the great circle between 
a point off the American coast, in latitude 42° N., longitude 
50° W., and Fastnet Rock off the Irish coast, in latitude 51° 24' 
N., longitude 9° 36' W. Required the initial course, the dis- 
tance between these points, and the latitude and longitude of 
the vertex. 

In the triangle CPC (Fig. 20) let O represent the initial point, C 
P Fastnet, and P the pole. 



c' = 90°-42° 0'=48° 0' 

= colatitude of C. 
c =90°- 51° 24'= 38° 36' 

= colatitude of O r . 
\ d = P=50°- 9° 36' = 40° 24' 

= difference of longitude of C and C". 




Fig. 20. 

tan£(C' + 0) 



To find the initial course : 
_ cos j (o f — c) 



cos I (c' -f c) 



C0t£A d , 



tani(O'-O) 



' — n\ = sinHc f -c) 



cot § \ 



W-c) ■ 
Uc' + c) 



4° 42' 
43° 18' 
20° 12' 



£(C" + C r ) = 74°58', 
l\C- C) = 17° 59', 

.-. C = 56°59', or N. 56°59'E. 



sin?(c' + c) 

log cos = 9.99854 

cologcos = 0.13800 

log cot = 10.43424 

log tan =10.57078 



[10] 



log sin - 8.91349 

cologsin= 0.16379 

log cot = 10.43424 



log tan = 9.51152 



GREAT CIRCLE SAILING. 



55 



To find the distance : 



cos i D 



008 * ft+ff sin U, 

cos*(G-f-0') 



J(c + c') =43° 18' 

£(C+C") = 74°58' 

£x d =20°12' 



£Z> = 14°21', 
. Z> = 28°42' 
= 1722 miles. 



log cos = 9.86200 

cologcos = 0.58606 

logsin = 9.53819 

log cos = 9.98625 



[ii] 



To find L of the vertex : 

sin PV = sin c' sin 0. 

c' = 48° (/, 
C=56°59', 

PV= 38° 33', log sin = 9.79458 
L of V= 90° - 38° 33' = 51° 27' N. 



[12] 

log sin = 9.87107 
log sin = 9.92351 



To find \ of the vertex : 

cotCPV = cose' tan 0. [13] 

log cose' = 9.82551 
log tan O =10.18721 

log cot CPP= 10.01272 
.-. CPF=44°10' 
AofC = 50° O'W. 



A of V= 5°50'W. 



Note. %(C+ C) = 74° 57' 42", using this value, D = 1730 miles. 
From Sandy Hook to the company's position 1088 miles, thence to Fast- 
net by great circle 1730 miles; total, 2818 miles. 



Exercise VIII. 

1. Find the elements (initial courses, distance, and latitude 
and longitude of the vertex) of the great circle track between 
the Lizard, in latitude 49° 58' N, longitude 5° 12' W., and the 
Bermuda Islands, in latitude 32° 18' N., longitude 64° 50' W. 

2. Find the elements of the great circle track between 
Boston (Minot's Ledge light-house), in latitude 42° 16' N., 
longitude 70° 46' W., and Cape Clear, in latitude 51° 26' N, 
longitude 9° 29' W. (Take I \ d = 30° 39'.) 

3. Find the elements of the great circle track between Van- 
couver Island, in latitude 50° N., longitude 128° W., and Hono- 
lulu, in latitude 21° 18' N., longitude 157° 52' W. 



56 NAVIGATION. 



4. Find the elements of the great circle track between Cape 
Clear, in latitude 51° 26' N., longitude 9° 29' W., and Sandy 
Hook, in latitude 40° N., longitude 74° W. 

5. Find the elements of the great circle track between 
Lizard Lights, in latitude 49° 58' N., longitude 5° 12' W., and 
Cape Frio, in latitude 23° S., longitude 42° W. 

6. Find the elements of the great circle track between Cape 
Frio and Cape Good Hope, in latitude 34° 20' S., longitude 
18° 30' E. (Beckon from the nearest pole.) 

7. Find the elements of the great circle track between 
Grand Port, Mauritius, in latitude 20° 24' S., longitude 57° 
47' E., and Perth, in latitude 32° 3' S., longitude 115° 45' E. 

8. Find the elements of the great circle track between A, 
in latitude 16° 38' N., longitude 70° 55' W., and B, in latitude 
48° 2' N, longitude 4° 35' W. 

9. A ship sails from A, in latitude 40° S., longitude 148° 30' 
E., to B, in latitude 12° 4' S., longitude 77° 14' W. Compare 
the great circle and the rhumb-line between A and B. 



CHAPTEE III. 

NAUTICAL ASTRONOMY.* 



§ 20. The Observed Altitude of a Heavenly Body. 

The Sextant. 

The Sextant is an instrument used by navigators for deter- 
mining the angular distance between two heavenly bodies, or 
the angular altitude of a heavenly body above the horizon. 



This instrument is shown in outline in Fig. 21. It consists 
of a frame ABO in the form of a sector whose arc AB is 
60° in extent, but divided into 120 equal parts reckoned as 

* For definitions see Spher. Trig., gg 63 and 64. 



58 NAVIGATION. 



degrees. CE is a movable radius which carries at C a mir- 
ror called the index-glass, whose plane is perpendicular to 
the .plane of the sextant; and at E an index which moves 
over the arc AB. At D is fixed the horizon-glass, also per- 
pendicular to the plane of the sextant, and parallel to CA ; 
that is, parallel to the index- glass when the index is at (A)'. 
Only a part of the horizon-glass is silvered. F is a telescope 
fixed to CA. 

To determine the angular distance between two heavenly- 
bodies S and S', the sextant is held in the hand so that a ray 
of light from S' passes through the unsilvered part of the 
horizon-glass, enters the telescope, and is seen as if by direct 
vision. • The arm CE is then moved so that a ray of light 
from S, being reflected by the mirrors at C and D, enters the 
telescope, taking the path SCDF. The movement of CE 
is continued until the image of S appears to be in con- 
tact with 8\ when the arc AE gives the angular distance 
required. 

For, it is a principle in optics that, " when a plane mirror is 
rotated in the plane of incidence, the direction of the reflected 
ray is changed by double the angle through which the mirror 
is turned."* Hence, the angular distance between the two 
bodies, which is the angle between the two lines SC and S'E, 
is double the angle ACE, or double the arc AE. But since 
half-degrees on AB are marked as degrees, AE will be the 
reading required*. 

If S' is the point on the sea horizon directly under S, the 
arc AE will give the observed altitude of S. 

The Quintant has an arc of 72°, and is therefore capable of 
measuring angles up to 144°. In other respects it is like the 
sextant. 

The Octant (commonly, but improperly, called the quadrant) 
has an arc of 45°, and is therefore capable of measuring angles 
up to 90°. In general construction it is similar to the sextant, 

* Deschanel's Natural Philosophy, page 892. 



CORRECTIONS OP THE OBSERVED ALTITUDE. 59 

for which it is a cheap substitute. It is much more easily 
handled, and yet sufficiently accurate for many purposes.* 

In Fig. 21, the arc AB, extending from zero to the left, is 
called the arc proper; the arc to the right of zero is called 
the arc of excess. The index is on the arc or off the arc 
according as it is on the arc proper or on the arc of excess. 

When the index-glass is parallel to the horizon-glass, the 
index should be at 0°. If it is not, the arc between the zero- 
point and the index is called the Index Error. The index 
correction is — when the index is on the arc, and -j- when the 
index is o^the arc. 

§ 21. Corrections of the Observed Altitude. 
Refraction. 

Let S (Fig. 22) represent the true position of a star, the 
position of the observer on the surface of the earth, and PP 
the upper limit of the atmosphere. 

According to the principle of refraction, a ray of light from 
B meeting the atmosphere at P will be bent toward the per- 
pendicular to the arc PP' at P. As the density of the 
atmosphere increases from P to 0, PO will be a curve, and 
the star will be seen in the direction OS' tangent to PO 
at 0. 

Astronomical refraction is the difference of direction of the 
two lines PS and OS'. 

The effect of refraction is to increase the apparent altitude 
of a celestial object, and is greatest when the body is near the 
horizon, and decreases if the altitude of the body increases. 

* The octant is fully equal to every-day work in the broad ocean ; for exam- 
ple, during the winter months in the North Atlantic. The delicate exactness 
of the sextant is quite thrown away when one can get only flying shots at the 
horizon, from the crest of a 60-feet wave. Showers of salt spray, with the 
chance of an occasional knock, certainly seem less suited to the sextant than 
to its hardier and more humble relative. — Captain Lechy, in '"Wrinkles' in 
Practical Navigation," p. 38. 



60 



NAVIGATION. 



Table XIII. gives the correction for refraction correspond- 
ing to every apparent altitude from 0° to 90°. This correction 
must evidently be subtracted from the apparent altitude. 




Fig. 22. 

The refraction varies slightly according to the condition of 
the atmosphere, but this variation is not taken into account in 
this chapter. 

Dip of the Horizon. 

Let BH" (Fig. 23) represent the level of the sea, the posi- 
tion of the observer at the height OB above the sea-level, C 
z the centre of the earth, 

and H" the farthest 
point on the surface 
of the earth visible 
from 0. On account 
of refraction, the vis- 
ual ray H"0 will be 
a curve concave to- 
ward C, and H n will 
be seen in the direc- 
Fig - 23 ' tion of OH 1 tangent 

to H"0 at 0. Let OH, perpendicular to OC, represent the 




PARALLAX. 



61 



true horizon, from which the apparent altitude of the body is 
estimated. 

The dip = angle HOW. 

The dip is very nearly proportional to the square root of 
the height of the observer above the sea-level. Its effect evi- 
dently is to increase the observed altitude of a heavenly body ; 
hence, the correction for dip must be subtracted from the 
observed altitude. 

This element is given in Table XV. 



Parallax. 

The difference between the positions of a heavenly body as 
seen from the centre of the earth and a point on its surface at 
the same time, is called geocentric parallax. 




Fig. 24. 



Let 8 (Fig. 24) represent the body, the position of the 
observer, C the centre of the earth, and Z the zenith of the 
observer. 

The geocentric parallax is the angle 

8=Z08-ZC8 



62 NAVIGATION. 



ZOS is the apparent zenith distance, and ZCS the geo- 
centric or true zenith distance. It is evident that geocentric 
parallax decreases if the altitude of the body increases. 

Let P = the parallax in altitude, 

Z = ZOS the apparent zenith distance, 
P= CO, 
D = CS. 



In the triangle COS, 



sin S sin COS sin P 
or 



CO CS ' P JD 

tt 7-, P sin if , N 

Hence, smjT= — — (a) 

If the body be in the horizon as at H, the angle H'\s called 
the horizontal parallax = P'. In this case, the angle Z= 90°, 
and the preceding formula becomes 

-m P 

Substituting sinP' for — in formula (a), and putting h = 

SOH, whence Z= 90° - h, 

sin P = sin P ' cos h. 

Since P and P ' are very small, the sines are nearly propor- 
tional to the angles ; hence, 

P=P'cosh.' 

The sun's mean horizontal parallax is about 8.75", whence 
the sun's parallax in altitude may be computed. 

This element is given in Table XIV. 

The correction for parallax must evidently be added to the 
apparent altitude. 

The parallax of the fixed stars is practically 0. 



THE SUN S APPARENT SEMI-DIAMETER. 



63 



The Suns Apparent Semi- Diameter. 

Let O (Fig. 25) represent the centre of the earth, C the 
centre of the sun, and OP a tangent to the surface of the sun. 



diameter of the sun as viewed 
from 0. 

Let CC'= D, 
00= P, 
P0'= r, 
angle POO' = S = the sun's 
semi-diameter. 

In the right triangle 0P0' 
r 



S = 




I) 



Fig. 25. 



The formula for horizontal parallax is 
sinP' 



P 7-v P 



whence, by substitution, 



sin S= — sin P', or S=^-P. 
P P 

nearly, since S and P' are very small. 

Hence, the semi-diameter of the sun may be found from its 
horizontal parallax. 

Note. The semi-diameter of a heavenly body, as viewed from the 
centre of the earth, is called the horizontal semi- diameter. The semi- 
diameter is sensibly the same whether viewed from C or 0, provided the 
body be in the horizon, since in this position CO' and OC are nearly 
equal. As the altitude of the body increases, OC decreases, and the 
semi-diameter as viewed from increases ; this augmentation of the semi- 
diameter is sensible in the case of the moon, but not in the case of the 



The sun's semi-diameter is given for every day in the year 
in Table XII. It is greatest when the sun is nearest the earth, 



64 NAVIGATION. 



about January 1 ; and least when the sun is farthest from the 
earth, about July 1. 

In order to find the true altitude of the sun's centre, the 
semi-diameter is added to the apparent altitude of the lower 
limb, and subtracted from the apparent altitude of the upper 
limb. 

Applying the Corrections. 

The observed altitude of the limb is given by the sextant. 

The index error being applied, the result is the observed 
altitude above the sea horizon. 

The correction for dip being applied, the result is the appar- 
ent altitude of the limb, referred to the true horizon. 

The true altitude of the body is the altitude of its centre as 
viewed from the centre of the earth, and is obtained by apply- 
ing the corrections for refraction, parallax, and semi-diameter 
to the apparent altitude of the limb. 

The fixed stars have neither parallax nor semi-diameter. 

The following symbols will be used in this chapter. 

O the sun. (3 the sun's upper limb. 

-0-the sun's centre. Q the sun's lower limb. 

Example 1. 1882, July 5, the observed altitude of a star 
was 18° 20' 30"; index correction, +1' 17"; height of eye, 18 ft. 
Find the true altitude. 

Solution I. 

Observed altitude 18° 20' 30" 

Index correction + 1' 17" 

18° 21' 47" 
Dip (Table XV.) -4' 9" 

Apparent altitude 18° 17' 38" 

Refraction (Table XIII.) -2' 54" 

True altitude 18°14'44" 






APPLYING THE CORRECTIONS. 



65 



Solution II. 

Observed altitude 18° 20' 30" r Index correction + V 17" 

Correction - 5' 46" j Dip . . . . -4' 9" 

I Refraction . . -2' 54" 

True altitude 18° 14' 44" 

The corrections are properly made in a certain order in the first solu- 
tion ; in the second solution the algebraic sum of the corrections is 
applied to the observed altitude. It is evident that the correction for 
refraction might differ by the two methods, since in one case it is based 
on an altitude of 18° 17' 38", and in the other on an altitude of 18° 2(X 
30" ; but in any case the difference would be so slight as to be practically 
unimportant in sea observations. The second method is commonly used 
in practice. 

Example 2. 1882, Aug. 9, the mean of observed altitudes 
of the sun's lower limb was 26° 0' 2"; index correction, — 0' 17"; 
height of eye, 18 ft. Find the sun's true altitude. 

Solution I. 

Observed altitude Q 26° 0' 2" 

Index correction — / 17" 

25° 59' 45" 
Dip -4' 9" 

Apparent altitude Q 25° 55' 36" 

Refraction— V 59" ) v*\Vi 

Parallax + C 8" J 

True altitude Q 25° 53' 45" 

Semi-diameter (Table XII.) +15' 49" 

True altitude -0" 26° 9' 34" 



Observed altitude Q 



Correction 



Solution II 
. 26° 0' 2" 



Semi-diameter . 

Parallax . . . 

+ 9' 32" \ Index correction 

I Dip .... 

, [ Refraction . . 



+ 15' 49" 
+ V 8" 

- (yi7" 

- 4' 9" 

- 1'59" 



True altitude -e- 26° 9' 34" 



66 NAVIGATION. 






Exercise IX 






From the following data compute the true altitude 


: 


1. 1882, April 1, obs. alt. star 25° 6' 10", 


mdex cor. +1'15", 


eye 17 ft. 


2. 1883, July 25, obs. alt. star 15°20 / 25 // , 


mdex cor. -2' 20", 


eye 16 ft. 


3. 1882, Jan. 16, obs. alt. Q 


18° 17' 30", 


ndex cor. +& 18", 


eye 18 ft. 


4. 1882, Nov. 4, obs. alt. Q 


30° 12' 40", 


index cor. 0' 0", 


eye 19 ft. 


5. 1882, Oct. 10, obs. alt. Q 


56°25 / 20 // , 


index cor. — 1'20", 


eye 17 ft. 


6. 1882, Aug. 7, obs. alt. Q 


60° 10' 10", 


index cor. -2' 15", 


eye 18 ft. 


7. 1882, Feb. 9, obs. alt. O 


31° 24' 35", 


index cor. 0' 0", 


eye 18 ft. 


8. 1882, Mar. 1, obs. alt. 


26°17 / 20", 


index cor. +2' 15", 


eye 18 ft. 


9. 1882, Junel8, obs. alt. (D 


20° 35' 30", 


mdex cor. +0' 18", 


eye 16 ft. 


10. 1882, June 12, obs. alt. 


36° 12' 10", 


index cor. +0' 25", 


eye 20 ft. 



§22. Time. 
Apparent Solar Time is measured by the daily motion of the 



An Apparent Solar Day is the interval of time between two 
successive transits of the sun over the same meridian. This 
interval is not always the same, for two reasons : 

1. The sun moves in the ecliptic which is oblique to the 
equator. 

2. The rate of the sun's motion in the ecliptic varies, being 
greatest when the earth is at perihelion, and least when the 
earth is at aphelion. 

Hence, a clock or a chronometer cannot be regulated to fol- 
low the true sun. 

The Mean Sun is an imaginary sun supposed to move in the 
equator at a uniform rate equal to the mean rate of the true 
sun. 

Mean Solar Time is the hour angle of the mean sun. 

Apparent Solar Time is the hour angle of the true sun. 

The Equation of Time is the difference between mean time 
and apparent time. It never exceeds about 16 minutes. 

Siderial Time is measured by the apparent daily motion of 
the fixed stars. It is the hour angle of the vernal equinox. 



TIME. 



67 



The Siderial Day is the interval between two successive tran- 
sits of the vernal equinox over the same meridian. It is about 
3 min. 56 sec. shorter than the mean solar day, and is divided 
into 24 hours numbered from to 24. 

The Civil Day is the interval of time between two successive 
midnights. It is divided into two periods of 12 hours each, 
the first of which is marked A.M., and the second p.m. 

The Astronomical Day begins at noon of the civil day of the 
same date. It is divided into 24 hours numbered from 
to 24. 

To convert civil into astronomical time : 

If the given time be a.m., deduct 1 day, add 12 hours, and 
omit the a.m. ; if the given time be p.m., simply omit the p.m. 

To convert astronomical into civil time : 

If the hours of the given time be less than 12, simply affix 
p.m. ; if the given hours be 12 or more, add 1 day, subtract 12 
hours, and affix A.M. 

Exercise X. 





Astronomical Time, 


Civil Time. 


1. 


d. h. m. 

1882 July 8 7 6 


8. 

10 


Required. 




2. 


1882 Mar. 7 12 25 


30 


Required. 




3. 


1880 Jan. 1 18 10 


10 


Required. 




4. 


1879 Dec. 31 15 





Required. 




5. 


1883 Feb. 2 8 4 


30 


Required. 




6. 


Required. 




1882 July 1 11 8 


25 A.M. 


7. 


Required. 




1880 Mar. 2 11 56 


56 p.m. 


8. 


Required. 




1880 Aug. 31 10 8 


20 p.m. 


9. 


Required. 




1881 Sept. 1 12 


15 A.M. 


10. 


Required. 




1883 Jan. 1 10 41 


56 A.M. 



§ 23. Longitude and Time. 

Since the earth revolves on its axis once in 24 hours, the 
sun appears to move over the 360° of the equator in the same 
time. Hence, the longitude of any place is proportional to the 



68 NAVIGATION. 



time required by the sun to move over an arc of the equator 
included between the meridian from which longitude is reck- 
oned and the meridian of the place; and longitude may be 
expressed either in time or arc. 
Since 360° = 24 hours, 

15° = 1 hour, "J ( 1° = 4 minutes, 

15' = 1 minute, > or \ V =4 seconds, 

15" = 1 second, J I 1" = -^ second. 

Either of the above tables enables us to convert longitude 

in time to longitude in arc, and vice versa. 

§ 24. Greenwich Date. 

The Greenwich date is the time at Greenwich correspond- 
ing to the local time elsewhere. Greenwich time should 
always be expressed astronomically. 

To find the Greenwich date, having given the ship time and the 
longitude of the ship : 

1. Express the ship time astronomically. (§ 22.) 

2. Convert longitude in arc into longitude in time. (§ 23.) 

3. If the ship be in west longitude, add the longitude in 
time ; if the ship be in east longitude, subtract the longitude 
in time. 

The resulting Greenwich date will be expressed in apparent 
time or mean time according as the local time is apparent or 
mean. 

It is evident that local time can be found from Greenwich 
time by reversing these steps. 

Example 1. June 4, at 6 h. 10 m. p.m., apparent time, at 
ship in longitude 30° 15' E. Find Greenwich date. 

d. h. m. 

Ship date = June 4 6 10 

Longitude in time . . . . _ = — 2 1 

Greenwich date (apparent time) = June 4 4 9 

1 5)30° 15' = longitude in arc. 
2h. 1 m. = longitude in time. 



THE CHRONOMETER. 



69 



Example 2. August 21, at 8h. 30 m. 10 s. a.m., mean time, 
at ship in longitude 90° 30' W. Find Greenwich date. 

d. h. m. 8. 

Ship date = Aug. 20 20 30 10 

Longitude in time . . . . = +620 

Greenwich date (mean time) . = Aug. 21 2 32 10 



sun on the meridian of the 
Find the Greenwich date. 

d. 



Example 3. March 20, the 
place in longitude 160° 15' W. 

h. m. s. 

Ship date ...... . = Mar. 20 000 

Longitude in time . . . . = + 10 41 

Greenwich date (apparent time) = Mar. 20 10 41 
Exercise XI. 



1. 

2. 
3. 
4. 
5. 


Local Civil Date. 


Longitude. 


Greenwich Date. 


May 
July 
Aug. 
Mar. 
Mar. 


d. h. m. s. 

4 6 12 15 p.m. 
31 11 12 30 a.m. 

1 2 10 15 a.m. 

2 10 20 p.m. 
25 11 8 p.m. 


170° 50 7 0"W. 

40° 20/ 0"W. 

80° 40' 45" W. 

50° 45' 0" E. 
100° 25' 30" W. 


Required. 
Required. 
Required. 
Required. 
Required. 


6. 

7. 
8. 
9. 




Required. 
Required. 
Required. 
Required. 


25° 7' 0"W. 

179° C 0"W. 

179° C 0"E. 

45' 0"E. 


Dec. 30 19 47 28 
July 4 23 51 
July 3 23 59 
May 19 19 40 20 


10. 




Required. 


2° 10' 0"E. 


Dec. 31 15 8 



§ 25. The Chronometer. 

A chronometer is a time-piece adapted to use on board ship. 

Unless otherwise specified, chronometers are supposed to be 
set to Greenwich mean time. 

The error of chronometer on mean time at any place is the 
difference between the mean time at that place and the time 
indicated by the chronometer. 

The Error of Chronometer on mean time at Greenwich is the dif- 



70 , NAVIGATION. 



ference between mean time at Greenwich and the time by chro- 
nometer. The error is fast or slow according as the chronom- 
eter is in advance of or behind the mean time at Greenwich. 

The Eate of Chronometer is the daily change in its error ; it 
is gaining or losing according as the chronometer is going too 
fast or too slow. 

To determine the error of chronometer on Greenwich mean 
time the longitude of the place must be known, whence the 
Greenwich mean time may be found (§ 24) ; this compared 
with the chronometer set to Greenwich time gives the error. 

The rate may be found by determining the error on different 
days, and noting the daily change, supposing it to be uniform. 

Example 1. February 12, the error of chronometer was 
9 m. 19.6 s. slow, and February 25, the error was 10 m. 30.2 s. 
slow. Find the rate. 

m. s. 

Feb. 12, slow 9 19.6 

Feb. 25, slow 10 30.2 

Change for 13 days 1 10.6 

60 

13)70.6 

Change for 1 day, or rate 5.4s. losing. 

Since the chronometer was slow, and the error increasing, the rate was 
losing. 

Example 2. 1882, April 8, p.m., an observation was made, 
when the time was April 8, 4 h. 10 m. 22 s. by chronometer, 
which was 18 m. 10 s. fast on Greenwich mean time on Jan. 1, 
1882, and 16 m. 17 s. fast on January 31. Find the Green- 
wich date by chronometer. 

m. s. 

Jan. 1, fast 18 10 

Jan. 31, fast 16 17 

Change for 30 days 1 53 

60 



3 0)113 

Rate 3.8s. losing. 



THE NAUTICAL ALMANAC. 71 

Since the chronometer was fast, and the error decreasing, the rate was 
losing. 

d. h. m. s. d. h. s. 

Time hy chronometer, April 8 4 10 22 Feb. 28 3^ 

Original error . . -16 17 Mar. 31 228.0 60 

8 3 53 5 Apri l 8 4 26.6 7 

Accumulated rate . +4 15.3 67 4 .38 .1 

Greenwich date . . April 8 3 57 20.3 67.17 .27 .07 

255.25 
255.25 s. = 4 m. 15.3 s. 

The error on Jan. 31 is 16 m. 17 s. fast, which we deduct. From Jan. 
31 to April 8d. 3h. 57 m. is nearly 67 d. 4h. 
= 67.17 d. The error for one day is 3.8 s.; 3 ' 8s - 



hence, for 67.17 d. the error is 3.8 s. X 67.17 228.0 = 3.8 x 60 

= 255.3 s. = 4 m. 15.3 s. This accumulated rate 26.6 =3.8x 7 

(4 m. 15.3 s.) must be added, and the result is the .38 = 3.8 X =1 

required Greenwich date. .27 = 3.8 X .07 

The method of multiplying 3.8 s. by 67.17, 255 25 = 3 8 x 67 17 
employed above (at the right of the page), will 
be made plain by the arrangement of the work shown in the margin. 



§ 26. The Nautical Almanac. 

The American and British Nautical Almanacs give, at equi- 
distant instants of Greenwich, time, the apparent right ascen- 
sions and declinations of the sun, moon, planets, and certain 
fixed stars ; the semi-diameters of the sun and moon ; the equa- 
tion of time; etc., etc. 

Table XII. contains the data from the Nautical Almanac 
which are necessary for the solution of the exercises in this 
chapter. 

To reduce the sun's declination by means of the hourly difference : 
Table XII. gives the sun's declination at apparent and mean 

noon, Greenwich time, and the hourly change in declination 

for every day in the year. 

If the declination at noon be required, it may be taken at 

once from the table. The declination at any other time may 

be found by simple interpolation ; that is, by assuming that 



72 NAVIGATION. 



the declination varies as the time. The column headed "Diff. 
for 1 hour," contains the average change for one hour. This 
multiplied by the number of hours gives the correction re- 
quired. 

Note. In Table XII. the difference of declination for 1 honr is 
marked + or — according as the snn is moving northward or southward, 
but in the following examples "Diff. for lh." is marked + or — accord- 
ing as the element to which it is to be applied is increasing or decreasing 
numerically. 

Example 1. Find the sun's declination for 1882, Jan. 7 d. 
10 h., apparent time, Greenwich date. 

O's dec. Jan. 7 d. h. = 22° 21' 5.6" S. Diff. for 1 h. = - 19.41" 

-3' 14.1" 10 



O's dec. Jan. 7d. 10 h. = 22° 17' 51.5" S. Diff. for 10 h. = - 194.1" 

= -3' 14.1" 
The table gives the declination for Jan. 7d. Oh. = 22° 21' 5.6" S., and 
the hourly difference = 19.41", which is marked — because the declina- 
tion is decreasing. For 10 h. the declination is 19.41" x 10 = 194.1" 
= 3' 14.1", which must be subtracted from the declination at noon. 

Example 2. Find the sun's declination for 1882, Feb. 12 d. 
17 h. 12 m. 10 s., mean time, Greenwich date. 

O's dec. Feb. 13 d. Oh. = 13° 17' 9.1" S. Diff. for lh. =-50.59" 

5' 43.9" 6.8 



O's dec. Feb. 12d. 17 h. Diff. for 6.8h. = 343.912" 

12 m. 10 s. = 13° 22' 52.0" S. ' =5' 43.9" 

In this example, the given date is nearer Feb. 13d. Oh. than Feb. 12 d. 
Oh. ; hence, we take from the table the declination for the former date. 
Feb. 13 d. h. - Feb. 12 d. 17 h. 12 m. 10 s. = 6 h. 47 m. 50 s. = 6.8 h., for 
which the difference is found to be 5' 43.9", which evidently must be 
added to the declination at noon. 

Example 3. 1882, May 5, apparent noon, longitude 57° W. ; 
required the sun's declination. 

d. h. m. 

Local time = May 5 

Longitude in time = 3 48 

Greenwich date = May 5 3 48 



THE NAUTICAL ALMANAC. 73 

O'sdec. May5d. Oh. =16° 18' 5.9" Diff. for lh. = + 42.66" 

+ 2' 42.1" 3.8 



0*8 dec. May 5 d. 3h. 48 m. = 16° 20' 48" Diff. for 3.8 h. = 162.108" 

= + 2' 42.1" 

To reduce the equation of time. 

Example 4. Greenwich date, mean time, 1882, March 2 d. 
3 h. 10 m. ; find the equation of time. 

m. s. 8. 

Eq. of time Mar. 2 d. h. = - 12 18.65 Diff. for 1 h. = - 0.524 

- 1.66 3.1f 

Eq. of time Mar. 2 d. 3 h. 10 m. = 12 16.99 Diff. for 3.1| h. = - 1.6593 

Table XII. gives the equation of time corresponding to mean time, Mar. 
2d. Oh. = 12m. 18.65s. The — sign indicates that the equation of time 
is to be subtracted from mean time to obtain apparent time. 

The difference for one hour is also found in the table and marked — , 
because the equation of time is decreasing. The difference for 3 h. 10 m. 
is found to be 1.66 s., which is subtracted from the equation of time at 
noon. 

This method is not strictly correct, because the change in declination 
is not uniform ; but it will be found accurate enough in most cases, and 
will be used in this chapter. 

Exercise XII. 

Find the sun's declination and the equation of time corre- 
sponding to the following Greenwich dates : 

d. h. m. s. 

1. 1882 Jan. 7 3 apparent time. 

2. 1882 Aug. 1 6 12 20 apparent time. 

3. 1882 May 5 10 25 apparent time. 

4. 1882 Aug. 7 15 12 apparent time. 

5. 1882 Dec. 4 6 18 apparent time. 

6. 1882 July 23 20 16 40 apparent time. 

7. 1882 Nov. 13 6 apparent time. 

8. 1882 Oct. 12 5 12 apparent time. 

9. 1882 June 7 3 18 mean time. 
10. 1882 Feb. 3 9 15 mean time. 



NAVIGATION. 



§ 27. To find the Hour Angle of a Heavenly Body. 

Let t = the hour angle of the heavenly body. 
L = the latitude of the observer. 
d = the decimation of the body. 
h = the true altitude of the body. 
p = 90° ± d = the polar distance of the body. 
S=l{L+>p+h). 
R = $(L+p-h) = $ (L ~jj + h) -h = 8—h. 

The general formula for the hour angle of a heavenly body 
in terms of its polar distance and altitude, and the latitude of 
the observer, is 

sin k t = ± [cos S sin E sec L esc p]*. [Spher. Trig. § 67.] 

Example 1. The true altitude of a heavenly body is 14° 
50' 42"; its declination ,16° 20' 5" S. ; and the latitude of the 
N. Find the hour angle. 






i -o 



h = 11° 50 42" 

L= 37° log sec 0.09765 

j p = 106°20' 5" log esc 0.01789 

2, ?= 158° 10'^' 

S= 79° 5' 24" log cos 9.27708 

B= 64° 14' 42" log sin 9.95456 



2 19.34718 



log sin |i = 9.67359 
.-. |* = 28° 8' 23" 

£ = 56° 16 46" 
= 3 h. 45 m. 7 s. 

§ 28. To find the Local Time. 

The local apparent time is the hour angle of the true sun. if 
the sun is west of the meridian ; if the sun is east of the meri- 
dian, the local apparent time is the difference between 24 hours 
and the hour angle. 



TO FIND THE LOCAL TIME. 



75 



Example 1. On board the U.S. Steamer Swatara, in lati- 
tude 18° 12' S., Greenwich mean time, 1882, Sept. lid. 14 h. 
55 m. 50 s. ; observed altitude of 0,27° 18' 25"; index error 
-f- 31" ; height of eye, 18 ft. Kequired the local mean time. 

Sept. 11 d. 14h. 55m. 50s. = 9 h. 4m. 10s. (= 9.07h.) before noon, Sept. 12. 
To reduce the sun's declination and find the polar distance : 

O'sdec. Sept. 12 d. Oh. = 4° 7'20.8"N. Diff.forlh. 
Change for 9.07 h. = 8' 40.3" 

O 's dec. Sept. lid. 14 h. 

55m.50s.= 4° 16' 1" N. 



Diff. for 9 h. 
Diff. for .07 h. 



-57.36" 

516.24" 
4.02'' 



90 c 



Diff. for 



520.26" 
07h.= 8' 40.3" 



p = 94°16' 1" 
To find the true altitude : 



Obs. alt. Q , 



26° 18' 20" 
+ 10' 29" 



semi-diam., + 15' 56" 
index error, + 31" 



parallax, 


+ 8" 


dip, 


- 4' 9" 


refraction, 


- 1'57" 



True altitude, 26° 28' 49" 
To reduce the equation of time : 



Eq.oftimeSept.12d. Oh. 
Change for 9.07 h. 

Eq.oftime Sept. lid. 14 h. 


= 3 48.62 
7.90 

= 3 40.7 


Diff.forlh. 

Diff. for 9 h. 
Diff. for .07 h. 

Diff. for 9.07 h. 


= +0.871 

= 7.839 
= .061 


55m. 50s. 


= 7.900 



To find the hour angle of the sun : 

sin \t = ± [cos #sin R sec L cscp]^. 
h = 26° 28' 49" 



L = 


18° 12' 0" 


log sec = 0.02229 


p = 


94° 16' 1" 


log esc = 0.00121 


IS = 


138° 56' 50" 




8 = 


69° 28' 25" 


log cos = 9.54486 


R = 


42° 59' 36" 


log sin = 9.83373 

2)19.40209 

log sin ^= 9.70105 
.-. £$ = 30° 9' 33" 
£ = 60° 19' 6" 



76 



NAVIGATION. 



t = 60° 19' 6" = 4 1 16 

24 



Sept. 11, 



19 58 44 
-3 41 



Local apparent astronomical time 
Equation of time 

Local mean astronomical time = Sept. 11, 19 55 
Local mean civil time = Sept. 12, 7 55 

If the time be A.M., apparent civil time = 12 h. — t. 



3 

3 A.M. 



Example 2. 1882, Aug. 23, p.m., at sea, in latitude 9° 37' 
N. Five observed altitudes of the Q were taken at the times 
(by the watch), standing opposite, viz. : 

Obs. alt. Q 21° 33' 30" 

21° 28' 10" 

21° 24' 00" 

21° 19' 10" 

21° 14' 30" 

Index error, ; height of eye, 18 ft. ; correction of watch by 
chronometer, — 5 h. 16 m. 49.3 s. Required the local time. 

The mean of the five observed altitudes is found by dividing their 
sum by 5, and the mean time (by a watch) of these observations is found 
in a similar manner. The Greenwich mean time is found to be Aug. 
22 d. 21 h. 26 m. 50 s., which is 2h. 33 m. 10 s. before noon, Aug. 23. 



2 42 58 


2 43 21.5 


2 43 39 


2 43 59.5 


2 44 18.5 



Obs. alt. Q . 










b. m. s. 


O's dec. 






21° 33' 30" 


2 42 58 


11° 24' 14.4" N. 


Diff. for 1 h. 


= + 50.97" 


21° 28' 10" 


2 43 21.5 


+ 2' 10.0" 


Diff. for 2 h. 


= 101.94" 


21° 24' 00" 


2 43 39 


11° 26' 24.4" N. 


Diff. for .5 h. 


= 25.48" 


21° 19' 10" 


2 43 59.5 


90° 


Diff. for .05 h. 


= 2.55" 


21° 14' 30" 


2 44 18.5 


78° 33' 36" 




129.97" 


119' 20" 


218 16.5 




Diff. for 2.55 h 


= 2' 10" 


21° 23' 52" 


2 43 39.3 










-5 16 49.3 




Semi-diam., 


+ 15' 52" 




9 26 50 




Par., 


+ 8" 




21 26 50 G. 


M.T. 


Dip, 


-4' 9" 




2 33 10 = 


2.55 h. 


Ref., 


- 2f 28" 



+ 9' 23" 



LATITUDE BY MERIDIAN OBSERVATION. 



77 



9' 23" 



h= 21° 33' 15" 
L = 9° 37' 00" 
p= 78° 33' 36" 

2 #=109° 43' 51" 
S= 54° 51' 56" 
i2= 33° 18' 41" 



log sec = 0.00615 
log esc = 0.00871 

log cos = 9.76004 

log sin = 9.73972 

2)19.51462 



log sin if = 9.75731 



h. m. 

4 39 



3 
+ 2 30 



Eq. of time. 

m. s. 

2 28.19 
1.65 



2 29.84 



Equation of time = 

Local mean time = 4 41 33 p.m. 



Diff.forlh. 

B. 

- 0.648 



1.296 
.324 
.032 

1.652 



2 
.5 

.05 



§29. Latitude by Meridian Observation op a 
Heavenly Body. 

Let HZH' (Fig. 25) represent the meridian of the observer, 
P the nearest pole, Z the zenith, HH { the horizon, EQ the 
equator, and S the heavenly body on the meridian. 




Fig. 25. 

QS = d = the declination of the body. 
HS= h = the true altitude of the body. 
SZ ■— z = 90° — h == the zenith distance of the body (named 
N. or S. according as the zenith is north or south of the body). 
QZ = PW = L = the latitude of the observer. 
QZ = SZ + QS, or L = % + d. 



78 NAVIGATION 



If the body be at S', 

QZ=S'Z- QS', or L = z-d. 

If the body be at S", 

QZ=QS"-ZJS", or L = d~z. 

If the body be at S"', 

QZ= QS W -ZS W , or L = d-z. 

Hence, to find the latitude of a place by meridian observa- 
tion of a heavenly body, 

If the declination and zenith distance of the body are of the 
same name, take their arithmetical sum ; if of different names, 
take their arithmetical difference and give it the name (iV 7 ". or S.) 
of the greater. 

(a) Latitude by meridian altitude of the sun. 

Example. 1882, Aug. 11, in longitude 92° 12' E., the 
observed meridian altitude of the sun's lower limb was 42° 
42'10"K; index correction, -2' 50"; height of the eye, 17 ft. ; 
required the latitude. 

Since the sun bears north, the zenith distance is south. 

d. h. m. s. 

Ship date, Aug. 11 

Longitude in time, 6 8 48 

Greenwich date, Aug. 10 17 51 12 

O's dec. Aug. lid. Oh. = 15° 14' 24" N. - 44.61 

4' 34" 6.15 



O's dec. Aug. lOd. 17h. 51m. 12s. = 15° 18' 58" N. 274.35 

Observed altitude Q, 42° 42' 10" 
Index correction, — 2' 50" 

42° 39' 20" 
Dip, -4' 2" 



Apparent altitude Q, 42° 35' 18" 
Refraction, - 1' 3" 



42° 34' 15" 
Parallax, + 7" 



True altitude Q, 42° 34' 22" 



LATITUDE BY MERIDIAN OBSERVATION. 



79 



42° 34' 22" 
+ 15' 50" 



42° 
90° 


50 7 12" 


47° 
15° 


9' 48" S. 
18' 58" N 



True altitude Q, 
Semi-diameter, 

True altitude, 



Zenith distance, 
Declination, 

Latitude, 31° 50' 50" S. 

The work may be arranged more compactly as follows: 

Long. 92° 12' E. = 6h. 8m. 48s. 

Index cor. 
Semi-diam 
Dip, 



q 42° 42' 10" 
+ 8' 2" 



Ref, 
Par., 



- 2' 50" 
+ 15' 50" 

- 4' 2" 

- V 3" 
+ 7" 



42° 50' 12" 
90° 

2 = 47° 9'48"S. 

Exercise XIII. 

From the following data find the latitude : 



O'sdec. 15° 14' 24" N. 
4' 34" 


44.61 
6.15 


d =15° 18' 58" N. 
z=47° 9'48"S. 

L =31° 50' 50" S. 


274.35 





Civil Date. 


Longitude. 


Obs. Merid. Alt. Q 


Index Cor. 


Eye. 


1. 


1882, Jan. 1. 


102° 41' W. 


59° 59' 50" S. 


+ 0' 50" 


15 


2. 


1882, Feb. 1. 


78° 14' E. 


78° 4'10"S. 


+ ty 55" 


12 


3. 


1882, Mar. 20. 


173° 18' W. 


89° 37' 0"N. 


+ 4' 32" 


18 


4. 


1882, April 1. 


87° 42' W. 


48° 42' 30" S. 


+ 1' 42" 


18 


5. 


1882, Sept. 1. 


97° 42' E. 


51° 4'50"S. 


-6' 0" 


23 


6. 


1882, Aug. 26. 


92° 3'E. 


35° 35' 20" N. 


+ 2' 17" 


12 


7. 


1882, May 16. 


45° 26' W. 


86° 34' 20" N. 


+ 4' 16" 


15 


8. 


1882, Mar. 20. 


174° 0'W. 


89° 56' 10" N. 


- 1' 15" 


15 


9. 


1882, June 1. 


44° 40' E. 


72° 14' 10" N. 


+ 3' 45" 


22 


,10. 


1882, Dec. 1. 


67° 56' E. 


18° 48' 10" S. 


-3' 6" 


18 


11. 


1882, Sept. 23. 


57° 45' E. 


84° 10' 50" N. 


- V 36" 


16 


12. 


1882, Sept. 23. 


119° 54' E. 


83° 46' 0"S. 


- 5' 30" 


18 


13. 


1882, Nov. 21. 


70° 20' E. 


80° 20' 0" N. 


- 2' 50" 


20 


14. 


1882, Dec. 31. 


123° 45' W. 


67° 8'10"S. 


+ / 9" 


13 


15. 


1882, Oct. 20. 


150° 2& W. 


49° 58' 50" N. 


+ V 10" 


19 


16. 


1882, June 1 . 


96° 17' E. 


75° 38' 15" N. 
Obs. a/t.Q. 


+ 0' 27" 


26 


17. 


1882, June 25. 


59°15'E. 


60° 23' 15" N. 


+ 2' 21" 


30 



80 



NAVIGATION. 



(b) Latitude by meridian altitude of a fixed star. 

Since the declination of a fixed star changes very slowly, it 
is sensibly the same for several successive days ; hence, the 
Greenwich date and the longitude are not required. 

Example. 1882, Dec. 21, the observed meridian altitude of 

the star Procyon (declination 5° 31' 16" N.) was 52° 51' 50" X.; 

index correction, —49"; height of eye, 21 ft. ; required the 

latitude. 

Obs. alt., 



52° 51' 50" X. 




( Index cor., 


- (y 49" 


-6' 2" pip, 


- 4' 29" 


(Ref, 


- V 41" 



True alt., 



52° 45' 4S" 
90 



Zenith dist., 37° 14' 12" S. 
Declination, 5° 31/ 16" X. 



Latitude, 



31° 42' 56" S. 



The following table gives the declinations necessary for 
Exercise XIV. 

Declixatioxs of Fixed Stars. 



Date. 


Star. 


Declination. 


1882, 


Jan. 


24. 


Aldebaran. 


16° l& 14" N. 


1882, 


Feb. 


1. 


Sirius. 


ICP 33' Z-J 


1882, 


Feb. 


12. 


Procyon. 


5° 31' 20" N. 


1882, 


Mar. 


19. 


Arcturus. 


io° 47' 33" ::. 


1882, 


Mar. 


31. 


Spica, 


io° sj :s." s. 


1882, 


April 


11. 


Sirius. 


16° 33' 0" S. 


1882, 


July 


6. 


Antares. 


26° 1(/ 13" S. 


1882, 


Aug. 


20. 


Altair. 


8° 33' 49" N. 


1882, 


Aug. 


20. 


# Centauri. 


59° 4# 30" S. 


1882. 


Sept, 


7. 


Arcturus. 


19° 47' 49" N. 


1882, 


Oct, 


7. 


Markab. 


14° 31' 46" N. 


1882, 


Nov. 


7. 


Fomalhaut, 


30° 11' 33" S. 


1882, 


Dec. 


i . 


a Arietis. 


22° 51' 33" X. 


1882, 


Dec. 


21. 


Procyon. 


5° 31' 16" N. 



LATITUDE BY EX-MERIDIAN ALTITUDE. 



81 



Exercise XIV. 
From the following data find the latitude : 





Civil Date. 


Star. 


Obs. Merid. Alt. 


Index Cor. 


Eye. 


1. 


1882, Jan. 24. 


Aldebaran. 


52° 36' 0"S. 


- O 7 23" 


20 


2. 


1882, Feb. 12. 


Procyon. 


77° 18' 10" S. 


+ Of 19" 


16 


3. 


1882, Mar. 19. 


Arcturus. 


36° 10 7 20" N. 


+ 2' 42" 


20 


4. 


1882, Aug. 20. 


Altair. 


66° 51' 10" N. 


+ (y 58" 


13 


5. 


1882, Nov. 7. 


Fomalhaut. 


59°40 / 0"N. 


+ V 12" 


23 


6. 


1882, Sept. 7. 


Arcturus. 


86° 35' 50" N. 


- 1' 10" 


12 


7. 


1882, Oct. 7. 


Markab. 


54° 10' 15" S. 





13 


8. 


1882, Aug. 20. 


# Centauri. 


59° 47' 13" S. 





25 


9. 


1882, Dec. 7. 


o Arietis. 


60° 29' 50" S. 


-2' 10" 


18 


10. 


1882, Feb. 1. 


Sirius. 


37° 50' 20" S. 


+ 1' 4" 


19 


11. 


1882, April 11. 


Sirius. 


61° 3'50"N. 





16 


12. 


1882, Mar. 31. 


Spica. 


52° 14/ o"N. 





19 


13. 


1882, July 6. 


Antares. 


70° 10' 30" N. 





21 



§ 30. Latitude by Ex-Meridian Altitude of the Sun, 
by Reduction to the Meridian. 

If it happens that the meridian altitude of the sun cannot 
be observed, owing to clouds or other causes, the altitude may 
be observed befcre or after noon and reduced to the meridian 
altitude. 

Towsons Method. 

Towson's Tables for the reduction of ex-meridian altitudes 
furnish a simple practical solution of the problem under con- 
sideration. 

Corresponding to the nearest hour angle (§ 27) and nearest 
declination (§ 26) of the sun at the time of observation, Tow- 
son's first table gives an Index Number and Augmentation I. 
The declination of the sun at noon is found by adding Aug- 
mentation I. to the declination at the time of observation. 

Corresponding to the Index Number (given by the first 
table) and the nearest true altitude of the sun at the time of 
observation, Towson's second table gives Augmentation II. 



82 



NAVIGATION. 



The true altitude of the sun at noon is found by adding Aug- 
mentation II. to the true altitude at the time of observation. 

Having the declination and altitude at noon, the latitude is 
found as in § 29. 

The latitude may be found from these tables by ex-meridian 
altitudes of a star or planet, but siderial time must be employed 
in finding the hour angle. 

If equal altitudes be observed before and after meridian pas- 
sage, the hour angle may be found by taking half of the elapsed 
time ; or, if the two altitudes differ only a few minutes (of arc) 
their mean may be reduced by employing half the elapsed 
time as the hour angle. 

It should be observed that Towson's method is independent 
of the latitude by account. 

The maximum altitude in Towson's Tables is 74°, and the 
hour angle corresponding cannot exceed 22 minutes. The 
greatest declination is 23° 20'. 

Towson's method will be illustrated by a single example. 

Example. 1882, Dec. 2, p.m., at ship, in longitude 4° 39' 
W. ; observed altitude O south of observer was 24° 14' 10" ; 
height of eye, 11 ft. ; time by watch, Dec. 2d. Oh. 50m. 58s. 
which had been found to be 19 m. 38 s. fast on apparent time 
at ship ; the difference of longitude made to the west was 21.3', 
after the error on apparent time was determined. Required 
the latitude. 



Time by watch, 
Watch fast, 



Diff. long. 2i.l, 



Dec. 



2 50 58 
-19 38 

2 31 20 
-1 25 



App. time at ship, Dec. 2 29 55 
Time from noon, 29 55 



App. time at ship, Dec. 2 29 55 
Longitude, 4° 39' W. +18 36 

Greenwich date, Dec. 2 48 31 
48 m 31s. = .81h. 



Dec. 2d. Oh., 21° 59' 53.1" 
18.7" 



Reduced declination, 
Augment. (Towson, Table I.), 

Meridian declination, 



22° 10' 29" 



23.13" 

.81" 



22° (yi2" 
+10' 17" 



_ 



LONGITUDE BY CHRONOMETER. 83 



Observed altitude Q, 24° 14/ 10" 

+ 11' 0" 



Semi-diam.,+16'16" 
Dip, - 3' 15" 

Parallax, + 0' 8" 
Kefraction, - 2f 9" 



True altitude, 24° 25' 10" 

Aug. (Towson, Table III., index 75), + 11' 56" 

Meridian altitude, 24° 37' 6" 

90° 



Zenith distance, 65° 22' 54" 

Meridian declination, 22° 10 7 29" 



Latitude, 43° 12' 25" 

§ 31. Longitude by Chronometer, from an Observed 
Altitude of the Sun* 

The difference of longitude of two places is equal to the dif- 
ference of time reduced to arc (§ 23). 

Hence, the longitude of a place from Greenwich may be 
determined by taking the difference between the local mean 
time and the Greenwich mean time, and reducing to arc. The 
longitude will be east or west according as the Greenwich time 
is less or greater than local time. 

The local mean time is found as in § 28. The Greenwich 
mean time is found by chronometer set to Greenwich time. 
(See § 25 ; also, the example of § 30.) 

* As an astronomical question, the determination of longitude resolves 
itself into the determination of the difference of time reckoned at the two meri- 
dians at the same absolute instant. For seamen, the only really practical 
methods of effecting this are : first by the chronometer, and secondly by lunars. 
These last, however, are rapidly dying out, and are mostly looked upon now 
as " fancy navigation." ... In the class of vessels most likely to need lunars 
(namely, those small craft which, for the sake of economy, carry but one chro- 
nometer), it is not likely than an expensive sextant or quintant will be found ; 
and if by chance it were, it is questionable whether the requisite expertness in 
observing and calculating would accompany it. Therefore, practically speaking, 
lunars are out of date. — Captain Lecky in" ' Wrinkles' in Practical Naviga- 
tion," p. 175. 



84 



NAVIGATION. 



Example. 1. 1882, February 10, a.m., at ship, latitude 50° 
48' N. ; observed altitude O 9° 10' 50" ; index correction, — 3' 
18" ; height of eye, 18 ft. ; time by chronometer, February 9 d. 
9h. 59 m. 25 s., which was 37 m. 58.8 s. fast for mean noon at 
Greenwich, December 20, 1881, and on January 10, 1882, was 
34m. 12s. fast for mean noon at Greenwich; from the time of 
observation until noon the ship has sailed on an S.W. course 
(true) 34 miles. Required the longitude at noon. 



December 20, chronometer fast, 37 58.8 
January 10, chronometer fast, 34 12.0 

Change in 21 days, 3 46.8 

60 



22-5.S 



T 3 )226.8 
21 J 7 )75.6 

I 10.8 losing 



324.0 

O 

60 J328.3 
5 2S.3 



30. 



The change in Id. is 226.8s. - 21 = 10.8s. ; the change in 30.4d. (the 
interval between noon January 10, and February 9d. lOh.) is 10.8s. x 
30.4 = 328.3s. = 5m. 28.3s. 



Time by chronometer, 
Original error, 



a. 
Feb. 9 



59 25 
34 12 



Accumulated rate, 
Greenwich mean time, 
Time, after noon 9h. 31m. 



9 9 25 13 
4-5 28 

9 9 30 41 

9.52 h. 



O's declination, 14° 36' 17. 7" S. 





7' 


39.6'' 


14° 
90° 


og/ 


38.1" 



= 104° 28' 38.1' 



-48.28" 
9.52" 

459.63" 
-' 39.6". 



Obs. alt. Q, 9° 10 7 50" f Semi-diam., + 16' 15" 
Correction, + 3' 10" \ Parallax, + 9" 



True alt.*©-, 9° 14' 0" 



Ind. cor., -3' 18" 
Dip, -4' 9" 

Refract, -5'47" 



LONGITUDE BY CHRONOMETER. 



85 



h= 9° 14' 0" 
L= 50° 48' 0" 
p = 104° 28' 38" 

2tf 164° 30' 38" 


log sec 0.19926 
log esc 0.01402 

log cos 9.12956 

log sin 9.98064 

2)19.32348 

sin$* = 9.66174 


Eq. of time. 

m. s. 

14 26.65 
+ .50 


s. 

+ 0.052 
9.520 


S = 82° 15' 19" 
E= 73° 1'19" 

log 


14 27.15 


.495 



Feb. 

Feb. 
Feb. 



.-. £f = 27°19'3" 
t = 54° 38' 6" 
= 3h. 38 m. 32s. Time before noon, Feb. 10. 

d. h. m. s. 

9 20 21 28 Local apparent astronomical time. 
+ 14 27 Equation of time. 

9 20 35 55 Local mean astronomical time. 
9 9 30 41 Greenwich mean time. 



11 5 14 Difference of time. 
166° 18' 30" E. long, at time of observation. 

Longitude at sights, 166° 19' E. Latitude at sights, 50° 48' N. 

Change until noon, 38' W. Change until noon, 24/ S. 

Longitude at noon, 165° 41' E. Latitude at noon, 50° 24' N. 

From the time of observation until noon the ship sailed S.W. 34 miles. 
The difference of latitude and of longitude corresponding to this course 
and distance are found by Middle Latitude Sailing to be 24' S. and 
38' W. respectively. 

Example 2. 1882, November 8, p.m., at sea, on board the 
U. S. Steamer Swatara, in latitude 30' S. ; five observed alti- 
tudes of the O were taken at the times (by watch) standing 
opposite, viz. : 



Obs. alt. O, 24 c 



51' 50" 

48' 50" 
44' 40" 
40' 20" 
37' 0" 



h. 
11 



3 
16 

33 
49 
64.5 

correction of watch 



Index error, +30" ; height of eye, 18 ft 
by chronometer, — 5h. 12m. 15.9s. Required the longitude. 



86 



NAVIGATION. 



24° 51' 50" 

48' 50" 

44/40" 
40' "20" 
37' 0" 



16 
33 
49 
64.5 



222' 40" 165.5 

24° 44' 32" 11 S 33.1 
- 5 12 15.9 



5 56 17.2 

= 5.94h. p.m. 



10' 35" 

h = 24° 55> 7" 

Z= 0°30' 0" log sec 0.00002 

p = 73° 17' 21" log esc 0.01874 

2^= 98° 42' 28" 
#=49° 21' 14" log cos 9.81383 
i2 = 24°26' 7" log sin 9.61665 
2)19.44924 



log sin |i= 9,72462 
.-. i = 4h. 16m. 



O's declination. 

16° 38' 20.6" S. 

+ 4' 18.4" 

16° 42' 39.0" 
90° 

73° 17' 21" 

Semi-diameter, 

Parallax, 

Index correction, 

Dip, 

Refraction, 

Eq. of time. 

m. s. 

16 7.21 
-1.13 



16 6.08 



+ 43.50" 
5.94" 

258.37" 
= 4' 18.4" 



+ 16' 12" 
+ 8" 

+ 30" 

- 4' 9" 

- 2' 6" 

+ 10 / 35" 



-.191 

5.94 

1.13 



16 s. Time after noon, Nov. 8. 



Nov. 8 4 16 16 Local apparent astronomical time. 

16 6 Equation of time. 

Nov. 8 4 10 Local mean astronomical time. 

Nov. 8 5 56 17 Greenwich mean time. 



1 56 7 Difference of time. 
29° 1' 45" TV. Longitude. 



Exercise XV. 

1. 1882, October 19, a.m., at sea, on board tbe U. S. Steamer 
Swatara, in latitude 33° 27' S. ; tbe observed altitude O, 28° 
22' 30" ; index correction, +30" ; height of eye, 18 ft. ; Green- 
wich mean time by cbronometer, October 18 d. 18 h. 28 m. 
38 s. Required tbe longitude. 



PROBLEMS. 87 



2. 1882, October 20, A.m., at sea, on board the Swatara, in 
latitude 31° 40' S. ; the observed altitude O, 35° 16' 10" ; index 
correction, +30" ; height of eye, 18 ft. ; Greenwich mean time 
by chronometer, October 19 d. 19 h. 11m. 24 s. Required the 
longitude. 

3. 1882, October 20, p.m., at sea, on board the /Swatara, in 
latitude 30° 55' S. ; the observed altitude O, 21° 42' 30" ; index 
correction, +29" ; height of eye, 18 ft. ; Greenwich mean time 
by chronometer, October 20 d. 3h. 35 m. 40 s. Required the 
longitude. 

4. 1882, October 21, a.m., at sea, on board the Swatara, in 
latitude 29° 35' S. ; the observed altitude O, 24° 26' 42"; index 
correction, +29 " ; height of eye, 18 ft. ; Greenwich mean time 
by chronometer, October 20 d. 18 h. 30 m. 39 s. Required the 
longitude. 

5. 1882, January 29, p.m., at ship, latitude 42° 26' N. ; 
observed altitude O, 13° 40'; index error, — 1'8"; height of 
eye, 16 ft.; time by chronometer, 29 d. 6h. 48m. 40s., which 
was slow 11m. 22.3 s. for mean noon at Greenwich, December 
1, 1881, and on January 1, 1882, was 8m. 7s. slow for Green- 
wich mean noon. Required the longitude. 

6. 1882, March 31, a.m., at ship, latitude 26° 9' N. ; observed 
altitude O, 29° 10' 20" ; height of eye, 26 ft. ; time by chro- 
nometer, 31 d. Oh. 4m. 50s., which was 58m. 58s. fast for mean 
noon -at Greenwich, November 20, 1881, and on December 31, 
1881, was 1 h. 2 m. 55.8 s. fast for mean time at Greenwich. 
Required the longitude. 

7. 1882, May 22, a.m., at ship, latitude 43° 25' N. ; observed 
altitude O, 32° 8'; index correction, +1' 28"; height of eye, 15 
ft.; time by chronometer, 21 d. 21 h. 6m. 10s., which was slow 
12.6 s. for mean noon at Greenwich, February 24, and on April 
1, was 2 m. 45 s. fast for mean noon at Greenwich. Required 
the longitude. 



88 NAVIGATION. 



8. 1882, August 24, a.m., at ship, latitude at noon, 37° 59' 
N.; observed altitude O, 37° 13' 30"; index correction, +2' 44"; 
height of eye, 18 ft. ; time by chronometer, August 23 d. 18 h. 
13 m. 24 s., which was lm. 5 s. fast for mean noon at Green- 
wich, August 1, and on August 10, was m. 42 s. slow for mean 
time at Greenwich ; course (true) since observation, N.N.W. ; 
distance, 22.4 miles. Required the longitude at noon. 

9. 1882, January 29, p.m., at ship, latitude 28° 45' N. ; 
observed altitude O, 17° 46' 30"; index correction, -3' 18"; 
height of eye, 16 ft. ; time by chronometer, January 28 d. 16 h. 
31m. 30s., which was lm. 16.5s. fast for Greenwich mean 
noon, December 17, 1881, and on January 1, 1882, was lm. 3s. 
slow for Greenwich mean time ; course (true) since noon, N.W. 
by "W. ; distance, 20 miles. Eequired the longitude at the 
time of observation, and also at noon. 

10. 1882, August 31, p.m., at ship, latitude 0° ; observed 
altitude O, 45° 5' 30" ; index correction, —2' 4"; height of 
eye, 15 ft.; time by chronometer, August 31 d. 9h. 11m. 28s., 
which was 5 m. 20 s. fast for mean noon at Greenwich, April 
15, and on June 16, was fast 2 m. 43 s. on mean time at Green- 
wich. Required the longitude. 

11. 1882, April 15, a.m., at ship, latitude 48° 52 f N.; 
observed altitude Q, 22° 18' ; index correction, —3' 54"; height 
of eye, 17 ft.; time by chronometer, April 14 d. 22 h. 30 m. 
42 s., which was m. 4 s. slow for mean noon at Greenwich, 
January 1, and on January 12, was fast m. 2 s. Required 
the longitude. 

12. 1882, August 28, p.m., at ship, latitude 5° S. ; observed 
altitude O, 38° ; index correction, +5' 27" ; height of eye, 21 
ft. ; time by chronometer, August 27 d. 22 h. 20m. 30 s., which 
was 10 m. s. slow for mean noon at Greenwich, February 19, 
and on May 30, was 2 m. 20 s. slow on mean noon at Green- 
wich. Required the longitude. 



_ 



PROBLEMS. 89 

13. 1882, September 22, a.m., at ship, on the equator, 
observed altitude (5, 17° 20' 40"; index correction, — 1' 9" ; 
height of eye, 20 ft. ; time by chronometer, September 22 d. 
4h. 59 m. 16 s., which was 15 s. slow for Greenwich mean noon, 
April 30, and on June 1, was 10.6 s. fast for mean time at 
Greenwich. Required the longitude. 

14. 1882, August 5, a.m., at ship, latitude at noon 30° 30' 
N. ; observed altitude O, 35° 6' ; height of eye, 15 ft. ; time by 
chronometer, 5 d. 8h. 39 m. 22 s., which was fast 29m. 32.4s. 
on Greenwich mean noon, July 8, and on July 20, was fast 
30m. s. on Greenwich mean noon ; course (true) till noon, W. ; 
distance, 48 miles. Required the longitude in at noon. 

15. 1882, November 12, a.m., at sea, on board the Swatara, 
in latitude 7° 10' N. ; four observed altitudes of the O were 
taken at the times (by watch) standing opposite, viz. : 

h. m. s. 

Obs. alt. O, 21° 8' 40" 2 55 48 

11' 50" 56 

14' 50" 56 13 

17' 30" 56 26.5 

Index correction, +31" ; height of eye, 18 ft. ; correction of 

watch by chronometer, — 5h. 12 m. 2.1s. Required the lon- 
gitude. 

16. 1882, November 13, a.m., at sea, on board the Swatara, 
in latitude 9° 30' N. ; five observed altitudes of the Q were 
taken at the times (by watch) standing opposite, viz. : 

Obs. alt. Q, 18° 58' 40" 2 59 2 

19° 1'20" 13 

3' 30" 28 

V 30" 45 

11' 0" 57.5 

Index correction, +32" ; height of eye, 18 ft. ; correction of 
watch by chronometer, — 5h. 11m. 58.4 s. Required the lon- 
gitude. 



90 NAVIGATION. 



17. 1882, November 17, a.m., at sea, on board the Swatara, 


in latitude 15° 35' N. ; five observed altitudes of the O were 


taken at the times (by watch) 


standing opposite, viz. : 


Obs. alt, Q, 23° 56' 0" 


h. m. s. 

4 12 31 


24° 0' 0" 


46 


4' 0" 


13 2.5 


6' 10" 


14 


10' 0" 


28.5 


Index correction, +31" ; height of eye, 18 ft. ; correction of 


watch by chronometer, — 5h. 


11m. 43.6 s. Required the Ion- 


gitude. 





18. 1882, November 18, A.M., at sea, on board the Swatara, 
in latitude 16° 25' N. ; five observed altitudes of the G) were 
taken at the times (by watch) standing opposite, viz. : 

Obs. alt. O, 18° 13' 30" 3 52 42 

16' 10" 53.5 

19' 20" 53 6.5 

22' 30" 23 

25' 30" 38 

Index correction, + 32" ; height of eye, 18 ft. ; correction of 
watch by chronometer, — 5h. 11m. 39.9 s. Required the lon- 
gitude. 

19. 1882, December 4, a.m., at sea, on board the Swatara, 
in latitude 36° 10' N. ; five observed altitudes of the O were 
taken at the times (by watch) standing opposite, viz. : 



Obs. alt. Q, 13° 0' 30" 






6 


m. s. 

27 14 


3' 10" 








27 29.5 


5' 40" 








27 49 


8' 50" 








28 5 


12' 0" 








28 23 


Index correction, +32"; 


h 


eight of ( 


sye, 18 ft. ; correction of 


watch by chronometer, — 5h. 


10 m, 


. 47.1s, Required the lon- 


gitude. 











DEVIATION BY TIME AZIMUTHS. 91 

20. 1882, December 4, p.m., at sea, on board the /Swatara, 
in latitude 36° 36' N. ; four observed altitudes of the O were 
taken at the times (by watch) standing opposite, viz. : 

h. m. s. 

Obs. alt. O, 16 c 



31' 10" 


1 7 26.5 




30' 0" 


35.5 




28' 30" 


46 




27' 20" 


56.5 




,+30"; hei 


ffht of eye, 18 ft. ; 


correction of 



watch by chronometer, — 5h. 10m. 46.1s. Required the lon- 
gitude. 

§ 32. Deviation by Time Azimuths. 

The Azimuth of a heavenly body is the arc of the horizon 
included between the north or south point and the vertical 
circle passing through the body. It is named by prefixing the 
letter (N. or S.) which indicates the point from which it is 
reckoned, and affixing the letter (E. or W.) which indicates 
upon which side of the meridian of the observer the body is 
situated. 

The True Azimuth is reckoned from the true north or south 
point. 

The Compass Azimuth is reckoned from the magnetic north 
or south point. 

When a heavenly body is on the horizon, the arc of the hori- 
zon between the body and the east or west point is called the 
Amplitude of the body. 

To find the compass azimuth of a heavenly body : 
The compass azimuth is observed directly by means of the 
Azimuth Compass. 

This instrument differs from the ordinary compass chiefly in 
having two sight vanes hinged to opposite points of the bowl. 

The standard compass is commonly also an azimuth compass. 

To observe the compass azimuth, the observer turns the 
compass so that, by looking through the narrow slit in one of 



92 NAVIGATION. 



the sight vanes, he can see the obje'ct, or the image of it (re- 
flected by a small mirror), bisected by the vertical wire of the 
other vane. A spring is then touched which holds the card in 
position until the observer reads off the division of the card, 
apparently in the prolongation of the wire of the vane. 

To find the true azimuth of a heavenly body : 

First Method ; by computation. The necessary data for time 
azimuths are : the declination (d), the hour angle (t), and the 
latitude (L). Let a be the true azimuth. 

The first and third formulas of Spher. Trig. § 68 apply. 

tan m = cot d cos t. 

tan a = sec ( L -f- ra) tan t sin m. 

Note. In these formulas w represents the polar distance of the foot 
of the perpendicular let fall from the body upon the meridian. 

Second Method; by inspection. Time azimuth tables have 
been prepared which give the true azimuth of the sun (and 
other heavenly bodies whose declinations lie between 23° N. 
and 23° S.) corresponding to the given latitude, declination, and 
apparent local time. This is the best practical method of find- 
ing true azimuths. 

To find the total error of the compass : 

The total error of the compass is equal to the difference be- 
tween the true and compass azimuths of a heavenly body, 
when they are both reckoned from the points of like name 
toward the east or toward the west ; it is equal to the sum of 
the true and compass azimuths when one is reckoned toward 
the east and the other toward the west. 

To name the total error of the compass, suppose the observer 
to look from the centre of the compass toward the two azi- 
muths ; then the error is east or west according as the true 
azimuth falls to the right or left of the compass azimuth. 

To find the deviation : 

If the total error of the compass and the variation (given by 
the chart) be of the same name (E. or W.), their difference will 






DEVIATION BY TIME AZIMUTHS. 93 

be the deviation ; if of different names, their sum will be the 
deviation. 

When the error exceeds the variation, the deviation is of 
the same name as the error ; but when the error is less than 
the variation, the deviation is of contrary name to the error. 

The deviation is thus determined for the point of the com- 
pass toward which the ship's head lies at the moment of obser- 
vation. By swinging the ship and repeating the observations, 
the deviation may be found for as many points of the compass 
as may be desired. Simple interpolation may be employed in 
finding the deviation for a point situated between two points 
for which the deviations are known. 



ANSWERS. 







Exercise I. 




1. 


S. 35°56'E. 


7. N. 56° 22' E. 


13. S. 42° 0' E. 


2. 


N. 77° 45' W. 


8. S. 73° 26' 


f ;W. 


14. N. 30°0'W. 


3. 


S.E. by E. 


9. S. 39° 15' E. 


15. N. 52°0'W. 


4. 


S. 79° 4'W. 


10. N.E. by E. 


16. S. 27°0'W. 


5. 


N. 82° 47' W. 


11. N.N.W. 




17. S. 17°0'W. 


6. 


S. 60°30'E. 


12. W, 




18. N. 17° 56' E, 






Exercise II. 






1. L ri 


= 45°26'N., 


p = 


405.8 W. 




2. L" 


= 2° 46' Si- 


p = 


405.6 E. 




3. X" 


- 0° 3'S., 


p = 


406.0 E. 




4. i" 


= 2° 12' N., 


p = 


52.1 E. 




5. IP 


= 36°57'N., 


p = 


113.1 W. 




6. 


= 61° 49', 


p = 


317.3. 




7. C 


= 25° 19', 


p = 


151.4. 




8. D 


= 118° 3', 


P = 


: 23.1 W. 




9. 7) 


= 199° 6', 


p = 


= 58.0 W. 




10. Z" 


= 37°28'N., 


D = 


: 58.0. 




11. L" 


= 17°54'N., 


Z> = 


= 72.0. 




12. C = 


= 25° 27', 


p = 


: 128.5. 




13. ■ 


= 22° 21', 


p = 


: 308.8. 




14. X" 


= 6° 3'N., 


P = 


95.4 E. 




15. X" 


= 14° 3'S., 


p = 


137.5 E. 




16. i": 


= 40°12'S., 


p = 


239.7 E. 




17. ■ 


= S. 53°22'W. 


,£ = 


: 360.4. 




18. (7 : 


= N.56° 6'E., 


Z>= 


■- 154.2. 




19. ■■ 


= N.75° 5'E., 


B = 


■ 147.7. 




20. 


= N.34°20'W. 


,r> = 


= 93.3. 



96 


NAVIGATION. 




Exercise III. 


1. 505.5 E. 5. 80.1 W. 


9. 172° 15' E. 


2. 125.0 E. 6. 166° 50' E. 


10. 12° 47' W. 


3. 109.1 W. 7. 30° 13' W 


11. 179° 11' E. 


4. 246.0 E. 8. 179° 48' E. 


12. 21° 18' W. ; 






N. 58° 40' E. 




Exercise IV. 


1. 


= E.N.E., 


D = 295.3. 


2. 


=N. 61°40'W., 


D = 210.7. 


3. 


C =S. 64°34'W., 


D = 209.6. 


4. 


X"=49°10'S., 


A" = 176° O'W. 


5. 


L" = 18° 52' N., 


X" = 175° 12' E. 


6. 


L"= 0°59'S., 


X" = 27°47'W. 


7. 


L" = 42° 11' N., 


X" = 65°46'W. 


8. 


L" = 41° 56' N., 


X" = 62°59'W. 


9. 


Z"=41°36'N., 


X" = 59° 12' W. 


10. 


A" = 57° 55' W., 


D= 61.5. 


11. 


X d = 3° 42', 


C = 87° 15'. 


12. 


=S. 57°19'E., 


D = 131.5. 


13. 


C = S. 71° 9' E., 


D = 510.7. 


14. 


= S. 53° 46' E., 


D = 289.3. 


15. 


(7 = S. 36°25'W., 


i) - 156.6. 


16. 


Z"=52°10'N., 


X" = 69° 13' W. 


17. 


Z" = 35° 14' N., 


X"=73° 2'W. 


18. 


i" == 42° 24' N., 


X" = 66° 15' W. 


19. 


i"=39°59'N., 


X" == 68° 35' W. 


20. 


Z"=53°20'N., 


X"= 1°25'W. 


21. 


£"=30° 4'N., 


X" = 67° 27' W. 


22. 


Z"=33° 8'N., 


X"=66°19'W. 


23. 


Z" = N. 64° 55' W., 


D = 675.5. 


24. 


L" = 25° 42' N., 
D = 775.8. 


C = S. 36° 27' W. 


' 







ANSWERS. 97 









Exercise V. 






1. 


=N.46°29'E., D = 140.9. 






2. 


G = S, 30° 23' E., Z> = 98.5. 






3. 


(7 == S. 28° 24' E., D = 261.5. 






4. 


C = N. 55° 13' E., D = 2160. 






5. 


== S. 43° 39' W., X> = 248.8. 






6. 


X" = 47° 10' N., A" = 32° 15' W. 






7. 


Z"=59° 3'K, A" = 2°43'E. 






8. 


Z" = 47° 20' N., a" = 13° 45' W. 






9. 


C = 33° 20', \ d = 13° 10'. 






10. 


A" = 23° 5' W, D = 1022. 






11. 


(7 =S. 51°59'W., D = 358.8. 






12. 


C = K 16° 2' E., D = 5599. 






13. 


C =S. 60°29'E., D = 7685. 






14. 


A d = 4° 48', B = 476.3. 




15. 


By Mercator's, L" = 37° 16' N, A" = 22° 47' W. ; 
Mid. Lat., Z" = 37° 16' N., A" = 22° 43' W. 








Exercise VI. 


1. 


L M * 


= 0°57'S.,^ = 21.4W. 5. Z d =0° 4'S.,jp = 25.0W. 


2. 


L d ~- 


= 1°37'S., p = 45.6W. 6. L d =0° 15'N.,^ = 36.5 W. 


3. 


W- 


= l°39 f S., p = 44.6W. 1.L d = 0° VS.,p= 6.8W. 


4. 


W- 


= o 45'N.,j9 = 37.1E. 8. i d =l° 11' S., p = 110.7 E. 








9. 2^=1° 2' S., i? = 119.5 E. 








Exercise VII. 






1. 


Z"=45°17'N., A" = 19°39'W. 






2. 


Z"=30°49'N., A" = 17°56'W. 






3. 


Z"=17°24'S., \" = 1°28'W. 






4. 


Z" = 46°51'N., A"= 48° 5'W. 






5. 


L" = 61° 34' N., A" = 149° 49' E. 






6. 


L" = 51° 29' S., A" = 176° 35' E. 






7. 


L" «= 36° 28' N., A" = 2° 10' W. 



98 NAVIGATION. 



Exercise VIII. 

1. Course from the Bermudas = N. 49° 32' E. ; 
Course from the Lizard = N. 88° 42' W. ; 
Distance = 2812 miles; L of V= 49° 59' N. ; 

AofF= 6°54'W. 

2. Course from Boston = N. 55° 24' E. ; 
Course from Cape Clear = N. 77° 42' W. ; 
Distance = 2500 miles ; L of V= 52° 28' N. ; 

A of V= 25° 2' W. 

3. Course from Honolulu = N. 32° 3' E. ; 
Course from Vancouver = S. 50° 17' W. ; 
Distance = 2228 miles; L of V= 60° 22' N. ; 

A of V= 80° 41' W. 

4. Course from Sandy Hook = N. 52° 50' E. ; 
Course from Cape Clear = N. 78° 18' W. ; 
Distance = 2698 miles ; L of V= 52° 23' N. ; 

A of V= 33° 42' W. 

5. Course from Cape Frio = N. 23° 2' E. ; 
Course from the Lizard = S. 34° 4' W. ; 
Distance = 4806 miles ; L of V= 68° 51' N. ; 

A of V= 34° 17' E. 

6. Course from Cape Good Hope = S. 85° 13' W. 
Course from Cape Frio = S. 63° 23' E. : 
Distance = 3208 miles ; L of V= 34° 37' S. ; 

AofF=10° 3'E. 

7. Course from Perth = S. 87° 45' E. ; 
Course from Grand Port == S. 64° 37' W. ; 
Distance = 3158 miles ; L of V= 32° 8' S. ; 

A of V= 111 29' E. 

8. Course from A = N. 43° 57' E. ; 
Course from B = S. 83° 51' W. ; 

Distance = 3714 miles ; L of V= 48° 19' N. ; 
A of 7= 3°39'E. 



ANSWERS. 



99 



9. By Great Circle : 

Course from A = S. 49° 
Course from B = S. 36° 
Distance = 6772 miles ; 

By Bhumb-line : 
Course from A = N. 76 c 
Course from B = S. 76 c 
Distance = 7432 miles. 



28' E. ; 
32' W. ; 
L of V= 54 c 
A of V= 95° 

' 58' E. ; 
' 58' W. ; 



24' S. 
26' E. 



1. 25° 1' 20". 

2. 15° 10' 41". 

3. 18° 27' 11". 

4. 30° 23' 4". 



1. 1882 July 8 7 6 

2. 1882 Mar, 8 25 

3. 1880 Jam 2 6 10 

4. 1880 Jan. 13 

5. 1883 Feb. 2 8 4 



EXERCISE IX. 

5. 56° 35' 28". 

6. 60° 19' 5". 

7. 31° 35' 13". 

Exercise X. 



8. 25° 57' 27". 

9. 20° 13' 41". 
10. 35° 51' 13". 



10 p.m. 6. 

30 a.m. 7. 

10 A.M. 8. 

a.m. 9. 

30 p.m. 10. 



1882 June 30 23 8 25. 

1880 Mar. 2 11 56 56. 

1880 Aug. 31 10 8 20. 

1881 Aug. 31 12 12 15. 

1882 Dec. 31 22 41 56. 



Exercise XI. 



d. h. m. s. 

1. May 4 17 35 35. 

2. July 31 1 53 50. 

3. July 31 19 32 58. 

4. Mar. 2 6 57 0. 

5. Mar. 25 17 49 42. 



6. Dec. 31 

7. July 4 

8. July 4 

9. May 20. 

1881. 

10. Jan. . 1 



h. m. s. 

6 7 a.m. 
11 55 p.m. 
11 55 p.m. 

7 43 20 a.m. 

3 16 40 a.m. 



Exercise XII. 



1. 22 c 

2. 17 c 

3. 16 c 



20' 7.4" S. ; 
55' 52.1" N. 
25' 30.3" N. 



6 38.85 to be added to app. time. 
6 3.66 to be added to app. time. 
3 30.82 to be subtracted from a.t. 





x^zi. v itr/iiiuii . 


4. 16° 13' 4.7" N. 


m. s. 

5 21.87 to be added to app. time. 


5. 22° 18' 47.4" S. ; 


9 28.67 to be subtracted from a.t. 


6. 19° 53' 5.6" N. ; 


6 14.27 to be added to app. time. 


7. 14° 32' 17.6" S. ; 


16 18.75 to be subtracted from a.t. 


8. 7° 32' 47.4" S. ; 


13 31.88 to be subtracted from a.t. 


9. 22° 47' 20.4" N. 


1 25.88 to be added to mean time. 


10. 16° 20' 41.1"; 


14 7.32 to be subtracted from m.t. 




Exercise XIII. 


1. 6° 49' 9"N. 


7. 16° 0' 55" N. 13. 29° 26' 44" S. 


2. 5° 23' 43" S. 


8. 0° 13' 46" N. 14. 0° 24' 44" S. 


3. 0. 


9. 4° 32' 48" N. 15. 50° 22' 22" S. 


4. 45° 47' 9" N. 


10. 49° 16' 24" N. 16. 7° 52' 4" N. 


5. 30° 30' 40" N. 


~\ 5° 43' 3"S. 17. 6° 31' 49" S. 


6. 43° 43' 32" S. 


i2. 6° 7'30"N. 




Exercise XIV. 


1. 53° 45' 44" N. 


6. 16° 19' 2" N. 10. 35° 40' 35" N. 


2. 18° 17' 0" N. 


7. 50° 28' 45" N. 11. 45° 34' 15" S. 


3. 34° 5' 7" S. 


8. 29° 30' 21" S. 12. 48° 24' 1" S. 


4. 14° 38' 0" S. 


9. 52° 31' 41" N. 13. 46° 4' 33" S. 


5. 60° 38' 37" S. 






Exercise XV. 


1. 16° 51' 30" E. 


10. 93° 51' 0"W. 


2. 13° 50' 45" E. 


11. 44° 30' 30" W. 


3. 12° 57' 30" E. 


12. 76° 4' 0"E. 


4. 10° 52' 30" E. 


13. 149° 18' 30" W. 


5. 49° 20' 15" W. 


14. 179° 17' 30" W. at sights. 


6. 41° 31' 0"W. 


179° 47' E. at noon. 


7. 20° 0'45"W. 


15. 34° 43' 30" W. 


8. 35° 27' 15" E. at 


sights. 16. 36° 50' 15" W. 


35° 16' E. at 


noon. 17. 45° 51' 0" W. 


9. 170° 54' 30" E. at 


sights. 18. 47° 5' 0"W. 


171° 13' E. at 


noon. 19. 74° 19' 15" W. 



20. 75° 16' 30" W. 



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